CAT 2025 Slot 1 QA Questions with Answers & Solutions

Q47MCQRatios, Proportions & Partnership

The ratio of the number of students in the morning shift and afternoon shift of a school was 13 : 9. After 21 students moved from the morning shift to the afternoon shift, this ratio became 19 : 14. Next, some new students joined the morning and afternoon shifts in the ratio 3 : 8 and then the ratio of the number of students in the morning shift and the afternoon shift became 5 : 4. The number of new students who joined is

110
121
88
99

Answer and solution

Answer: (D) 99

Let the shifts start with

13x

(morning) and

9x

(afternoon) students. After 21 students move,

\dfrac{13x - 21}{9x + 21} = \dfrac{19}{14}

, so

14(13x - 21) = 19(9x + 21)

, i.e.

182x - 294 = 171x + 399

. Then

11x = 693

and

x = 63

. The shifts now have

13 \times 63 - 21 = 798

and

9 \times 63 + 21 = 588

students. Let

3y

new students join the morning shift and

8y

the afternoon shift:

\dfrac{798 + 3y}{588 + 8y} = \dfrac{5}{4}

, so

3192 + 12y = 2940 + 40y

, giving

28y = 252

and

y = 9

. Check: the shifts become

825

and

660

, and

825 : 660 = 5 : 4

. New students

= 3y + 8y = 11y = 99

. Every option is a multiple of 11, so each could be split

3 : 8

; only

y = 9

gives

5 : 4

. For option A (110),

y = 10

gives

828 : 668

, and

4 \times 828 = 3312 \ne 5 \times 668 = 3340

. Hence, option D (99).

Q48MCQLogarithms

The number of distinct integers

n

for which

\log_{1/4}(n^2 - 7n + 11) > 0

is

0
1
Infinite
2

Answer and solution

Answer: (A) 0

The base $1/4$ lies between 0 and 1, so $\log_{1/4}(X) > 0$ exactly when $0 < X < 1$ . Both halves matter: the upper bound, and the requirement that the argument be positive at all.

Upper bound. $n^2 - 7n + 11 < 1 \Rightarrow n^2 - 7n + 10 < 0 \Rightarrow (n-2)(n-5) < 0$ , so $2 < n < 5$ . The only integers are $n = 3$ and $n = 4$ .

Positivity. $n^2 - 7n + 11$ has roots $\frac{7 \pm \sqrt{5}}{2}$ , about $2.38$ and $4.62$ , so the expression is positive only for $n < 2.38$ or $n > 4.62$ . This overlaps the first range only on $2 < n < 2.38$ and $4.62 < n < 5$ , which contain no integers, so it excludes both $n = 3$ and $n = 4$ .

Checking the two candidates directly: at $n = 3$ , $9 - 21 + 11 = -1$ ; at $n = 4$ , $16 - 28 + 11 = -1$ . Both are negative, so the logarithm is undefined at both and neither qualifies.

The two conditions have no integer in common, so the count is $\mathbf{0}$ — option A.

Q49MCQPolygons & Circles

In a circle with center C and radius

6\sqrt{2}

cm, PQ and SR are two parallel chords separated by one of the diameters. If

\angle PQC = 45^\circ

, and the ratio of the perpendicular distance of PQ and SR from C is 3:2, then the area, in sq. cm, of the quadrilateral PQRS is

$20(3+\sqrt{14})$
$4(3+\sqrt{14})$
$4(3\sqrt{2}+\sqrt{7})$
$20(3\sqrt{2}+\sqrt{7})$

Answer and solution

Answer: (A) $20(3+\sqrt{14})$

CP and CQ are radii, so triangle PCQ is isosceles. With

\angle PQC = \angle QPC = 45^\circ

, we get

\angle PCQ = 90^\circ

. Let M and N be the feet of the perpendiculars from C to PQ and SR; they bisect the chords. In right triangle CMQ the angle at Q is

45^\circ

, so

CM = MQ = \dfrac{CQ}{\sqrt{2}} = \dfrac{6\sqrt{2}}{\sqrt{2}} = 6

cm. Hence

PQ = 12

cm. The distances are in the ratio

3 : 2

, so

CN = 4

cm. Then

NR = \sqrt{72 - 16} = \sqrt{56} = 2\sqrt{14}

cm, and

SR = 4\sqrt{14}

cm. A diameter separates the chords, so they lie on opposite sides of C, and the distance between them is

6 + 4 = 10

cm. PQRS is a trapezium with parallel sides PQ and SR: Area

= \dfrac{1}{2}(12 + 4\sqrt{14}) \times 10 = 60 + 20\sqrt{14} = 20(3 + \sqrt{14})

sq. cm. Option B,

4(3 + \sqrt{14})

, comes from placing both chords on the same side of C, which gives a height of

6 - 4 = 2

cm instead of 10 cm. Hence, option A (

20(3+\sqrt{14})

).

Q50MCQAverages, Mixtures & Alligations

A container holds 200 litres of a solution of acid and water, having 30% acid by volume. Atul replaces 20% of this solution with water, then replaces 10% of the resulting solution with acid, and finally replaces 15% of the solution thus obtained, with water. The percentage of acid by volume in the final solution obtained after these three replacements, is nearest to

27
25
29
23

Answer and solution

Answer: (A) 27

The container starts with 200 L, of which

30\%

, or 60 L, is acid. Each step removes some solution and adds back the same volume, so the total stays 200 L. Step 1: removing

20\%

of the solution removes

20\%

of the acid, leaving

0.8 \times 60 = 48

L. Water is added, so the acid stays 48 L (

24\%

). Step 2: removing 20 L (

10\%

) leaves

0.9 \times 48 = 43.2

L of acid; then 20 L of acid is added, giving

63.2

L (

31.6\%

). Step 3: removing 30 L (

15\%

) leaves

0.85 \times 63.2 = 53.72

L of acid, and water is added. Final acid percentage

= \dfrac{53.72}{200} \times 100 = 26.86\%

. This is

0.14

away from 27 but

1.86

away from 25 (option B), the next-closest option, and further still from 29 and 23. Hence, option A (27).

Q51MCQTime, Speed & Distance

Shruti travels a distance of 224 km in four parts for a total travel time of 3 hours. Her speeds in these four parts follow an arithmetic progression, and the corresponding time taken to cover these four parts follow another arithmetic progression. If she travels at a speed of 960 meters per minute for 30 minutes to cover the first part, then the distance, in meters, she travels in the fourth part is

76800
112000
86400
96000

Answer and solution

Answer: (C) 86400

Work in metres and minutes: the total distance is 224000 m and the total time 180 min. The first part takes 30 min at 960 m/min, covering

960 \times 30 = 28800

m. The times form an AP with first term 30 and sum 180:

\dfrac{4}{2}(60 + 3d) = 180

, so

d = 10

and the times are 30, 40, 50 and 60 min. Let the speeds be

960

,

960 + D

,

960 + 2D

and

960 + 3D

. Then

960(30) + (960 + D)(40) + (960 + 2D)(50) + (960 + 3D)(60) = 224000

172800 + 320D = 224000

, so

D = 160

. The speeds are 960, 1120, 1280 and 1440 m/min, giving distances 28800, 44800, 64000 and 86400 m, which add to 224000 m. The fourth part covers

1440 \times 60 = 86400

m. Option A (76800) is

1280 \times 60

: it uses the third part's speed with the fourth part's time. Hence, option C (86400).

Q52TITAInequalities & Modulus

The number of distinct pairs of integers

(x, y)

satisfying the inequalities

x > y \geq 3

and

x + y < 14

is

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 16

We need integers with

x > y \ge 3

and

x + y < 14

. Since

x \ge y + 1

, we get

2y + 1 \le x + y < 14

, so

2y < 13

and

y \le 6

. So

y

is 3, 4, 5 or 6. For each

y

,

x

runs over the integers with

y < x < 14 - y

:

y = 3

:

3 < x < 11

, so

x = 4, 5, \dots, 10

(7 pairs).

y = 4

:

4 < x < 10

, so

x = 5, 6, \dots, 9

(5 pairs).

y = 5

:

5 < x < 9

, so

x = 6, 7, 8

(3 pairs).

y = 6

:

6 < x < 8

, so

x = 7

(1 pair). Total

= 7 + 5 + 3 + 1 = 16

. The answer is 16.

Q53MCQFunctions & Graphs

Let

3 \leq x \leq 6

and

[x^2] = [x]^2

, where

[x]

is the greatest integer not exceeding

x

. If set

S

represents all feasible values of

x

, then a possible subset of

S

is

$\left(3, \sqrt{10}\right) \cup \left[5, \sqrt{26}\right) \cup \{6\}$
$\left[3, \sqrt{10}\right] \cup \left[5, \sqrt{26}\right]$
$\left[3, \sqrt{10}\right] \cup \left[4, \sqrt{17}\right] \cup \{6\}$
$\left(4, \sqrt{18}\right) \cup \left[5, \sqrt{27}\right) \cup \{6\}$

Answer and solution

Answer: (A) $\left(3, \sqrt{10}\right) \cup \left[5, \sqrt{26}\right) \cup \{6\}$

Take each value of

[x]

in turn.

3 \le x < 4

: we need

[x^2] = 9

, i.e.

9 \le x^2 < 10

, so

3 \le x < \sqrt{10}

.

4 \le x < 5

: we need

16 \le x^2 < 17

, so

4 \le x < \sqrt{17}

.

5 \le x < 6

: we need

25 \le x^2 < 26

, so

5 \le x < \sqrt{26}

.

x = 6

:

[36] = 36

, which works. So

S = [3, \sqrt{10}) \cup [4, \sqrt{17}) \cup [5, \sqrt{26}) \cup \{6\}

. A:

(3, \sqrt{10}) \cup [5, \sqrt{26}) \cup \{6\}

lies inside

S

, so it is a subset. B and C both contain

\sqrt{10}

, which is not in

S

(

[10] = 10 \ne 9

). D contains numbers between

\sqrt{17}

and

\sqrt{18}

, which are not in

S

. The answer is A.

Q54TITAAverages, Mixtures & Alligations

Kamala divided her investment of Rs 100000 between stocks, bonds, and gold. Her investment in bonds was 25% of her investment in gold. With annual returns of 10%, 6%, 8% on stocks, bonds, and gold, respectively, she gained a total amount of Rs 8200 in one year. The amount, in rupees, that she gained from the bonds, was

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 900

Let investment in stocks be $S$ , bonds be $B$ , gold be $G$ .
$S + B + G = 100000$ and $B = 0.25G$ .
Total return = $0.10S + 0.06B + 0.08G = 8200$ .
Substitute $S = 100000 - 1.25G$ and $B = 0.25G$ :
$0.10(100000 - 1.25G) + 0.06(0.25G) + 0.08G = 8200$
$10000 - 0.125G + 0.015G + 0.08G = 8200$
$10000 - 0.03G = 8200 \Rightarrow 0.03G = 1800 \Rightarrow G = 60000$ .
Then $B = 0.25 \times 60000 = 15000$ .
Gain from bonds = $0.06 \times 15000 = \mathbf{900}$ rupees.

Q55MCQSimple & Compound Interest

At a certain simple rate of interest, a given sum amounts to Rs.13920 in 3 years, and to Rs 18960 in 6 years and 6 months. If the same given sum had been invested for 2 years at the same rate as before but with interest compounded every 6 months, then the total interest earned, in rupees, would have been nearest to

3221
3180
3150
3096

Answer and solution

Answer: (A) 3221

Let principal be $P$ , simple rate $R\%$ .
$A_1 = P(1 + 3R/100) = 13920$
$A_2 = P(1 + 6.5R/100) = 18960$
Difference in interest for 3.5 years = $18960 - 13920 = 5040$ .
Simple interest per year = $5040 / 3.5 = 1440$ .
Interest for 3 years = $3 \times 1440 = 4320$ .
Principal $P = 13920 - 4320 = 9600$ .
Rate $R = (1440 / 9600) \times 100 = 15\%$ .
If compounded semi-annually for 2 years at 15% p.a., rate per period is 7.5% and number of periods is 4.
Amount $A = 9600(1 + 0.075)^4 \approx 12820.5$ .
Total CI earned = $12820.5 - 9600 = 3220.5 \approx \mathbf{3221}$ .

Q56MCQPermutations & Combinations

A cafeteria offers 5 types of sandwiches. Moreover, for each type of sandwich, a customer can choose one of 4 breads and opt for either small or large sized sandwich. Optionally, the customer may also add up to 2 out of 6 available sauces. The number of different ways in which an order can be placed for a sandwich, is

840
800
880
600

Answer and solution

Answer: (C) 880

The customer chooses a sandwich type (5 ways), a bread (4 ways) and a size (2 ways). Sauces are optional, up to 2 of the 6: no sauce: 1 way; one sauce:

\binom{6}{1} = 6

ways; two sauces:

\binom{6}{2} = 15

ways. So there are

1 + 6 + 15 = 22

sauce choices. Number of orders

= 5 \times 4 \times 2 \times 22 = 880

. Option A (840) is

5 \times 4 \times 2 \times 21

: it leaves out the order with no sauce, which the word 'optionally' allows. Hence, option C (880).

Q57MCQTime & Work

Arun, Varun and Tarun, if working alone, can complete a task in 24, 21, and 15 days, respectively. They charge Rs 2160, Rs 2400, and Rs 2160 per day, respectively, even if they are employed for a partial day. On any given day, any of the workers may or may not be employed to work. If the task needs to be completed in 10 days or less, then the minimum possible amount, in rupees, required to be paid for the entire task is

47040
38880
34400
38400

Answer and solution

Answer: (D) 38400

Cost of the whole task if one person did it alone: Arun

24 \times 2160 = 51840

, Varun

21 \times 2400 = 50400

, Tarun

15 \times 2160 = 32400

. When a share of the work takes whole days, its cost is that share of the worker's whole-task cost. So give as much work as possible to the cheapest worker, Tarun, and the rest to the next cheapest, Varun. In 10 days Tarun does

\dfrac{10}{15} = \dfrac{2}{3}

of the task for

10 \times 2160 = 21600

. The remaining

\dfrac{1}{3}

takes Varun

\dfrac{21}{3} = 7

whole days, working alongside Tarun within the 10 days, for

7 \times 2400 = 16800

. Total

= 21600 + 16800 = 38400

rupees. Option B (38880) uses Arun for the last third instead:

\dfrac{24}{3} = 8

days cost

8 \times 2160 = 17280

, which is 480 more. Hence, option D (38400).

Q58TITAQuadratic & Polynomial Equations

The number of non-negative integer values of

k

for which the quadratic equation

x^2 - 5x + k = 0

has only integer roots, is

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 3

Let the integer roots be

\alpha

and

\beta

. Then

\alpha + \beta = 5

and

\alpha\beta = k

. We need

k \ge 0

. Both roots cannot be negative, since they add to 5. If one were negative, the other would exceed 5 and the product would be negative. So both roots are non-negative integers adding to 5. The unordered pairs are:

(0, 5)

, giving

k = 0

;

(1, 4)

, giving

k = 4

;

(2, 3)

, giving

k = 6

. Swapping the roots gives the same

k

. So

k

can be 0, 4 or 6: three values. The answer is 3.

Q59MCQCoordinate Geometry

The

(x, y)

coordinates of vertices P, Q and R of a parallelogram PQRS are

(-3, -2)

,

(1, -5)

and

(9, 1)

, respectively. If the diagonal SQ intersects the x-axis at

(a, 0)

, then the value of

a

is

$\dfrac{10}{3}$
$\dfrac{29}{9}$
$\dfrac{13}{4}$
$\dfrac{27}{7}$

Answer and solution

Answer: (B) $\dfrac{29}{9}$

The diagonals of a parallelogram bisect each other, so the midpoint M of PR is also the midpoint of SQ. Solution figure for question 59, CAT 2025 Slot 1

Solution figure for question 59, CAT 2025 Slot 1

Midpoint of PR

= \left(\dfrac{-3 + 9}{2}, \dfrac{-2 + 1}{2}\right) = \left(3, -\dfrac{1}{2}\right)

. If

S = (x, y)

, then

\dfrac{x + 1}{2} = 3

and

\dfrac{y - 5}{2} = -\dfrac{1}{2}

, so

S = (5, 4)

. Slope of SQ

= \dfrac{4 - (-5)}{5 - 1} = \dfrac{9}{4}

, so the line through

Q(1, -5)

is

y + 5 = \dfrac{9}{4}(x - 1)

. On the x-axis

y = 0

:

5 = \dfrac{9}{4}(a - 1)

, so

a - 1 = \dfrac{20}{9}

and

a = \dfrac{29}{9}

. The options are close together, so check the nearest one. Option C,

a = \dfrac{13}{4}

, gives

y = -5 + \dfrac{9}{4} \times \dfrac{9}{4} = \dfrac{1}{16}

on line SQ, not 0. Hence, option B (

\dfrac{29}{9}

).

Q60TITADigits & Base Systems

In a 3-digit number

N

, the digits are non-zero and distinct such that none of the digits is a perfect square, and only one of the digits is a prime number. Then, the number of factors of the minimum possible value of

N

is

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 6

The digits of $N$ are non-zero (1 to 9), distinct, and none are perfect squares.
Allowed digits (not 1, 4, 9) = $\{2, 3, 5, 6, 7, 8\}$ .
Exactly one digit must be prime. The primes in the set are $\{2, 3, 5, 7\}$ . The non-primes are $\{6, 8\}$ .
To form a 3-digit number, we need two non-primes and one prime. So the digits must be 6, 8, and one prime.
To minimize the number, we place the smallest digit at the hundreds place. The smallest prime is 2.
So the digits are 2, 6, 8. The minimum possible number is 268.
Prime factorization of 268 = $2^2 \times 67^1$ .
Number of factors = $(2+1)(1+1) = 3 \times 2 = \mathbf{6}$ .

Q61TITALinear Equations

Stocks A, B and C are priced at rupees 120, 90 and 150 per share, respectively. A trader holds a portfolio consisting of 10 shares of stock A, and 20 shares of stocks B and C put together. If the total value of her portfolio is rupees 3300, then the number of shares of stock B that she holds, is

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 15

Price of stocks A, B, C: 120, 90, 150. Shares of A = 10. Shares of B+C = 20.
Value of A shares = $10 \times 120 = 1200$ .
Total value = 3300. Value of B and C shares = $3300 - 1200 = 2100$ .
Let shares of B be $b$ and shares of C be $c$ . $b + c = 20 \Rightarrow c = 20 - b$ .
Value of B and C = $90b + 150(20 - b) = 2100$ .
$90b + 3000 - 150b = 2100$
$-60b = -900 \Rightarrow b = \mathbf{15}$ .

Q62MCQLinear Equations

If

a - 6b + 6c = 4

and

6a + 3b - 3c = 50

, where

a, b

and

c

are real numbers, the value of

2a + 3b - 3c

is

18
15
20
14

Answer and solution

Answer: (A) 18

Label the equations

a - 6b + 6c = 4

(1) and

6a + 3b - 3c = 50

(2). From (2),

3b - 3c = 50 - 6a

, so the required value is

2a + 3b - 3c = 2a + 50 - 6a = 50 - 4a

. We only need

a

. Doubling (2) gives

12a + 6b - 6c = 100

. Adding (1) cancels

b

and

c

:

13a = 104

, so

a = 8

. Then

2a + 3b - 3c = 50 - 32 = 18

. Check:

3b - 3c = 50 - 48 = 2

, and (1) gives

8 - 2 \times 2 = 4

, as required. Option D (14) is

2a - (3b - 3c) = 16 - 2

, a sign slip. Hence, option A (18).

Q63MCQQuadratic & Polynomial Equations

A value of

c

for which the minimum value of

f(x) = x^2 - 4cx + 8c

is greater than the maximum value of

g(x) = -x^2 + 3cx - 2c

, is

$-2$
$-\dfrac{1}{2}$
$2$
$\dfrac{1}{2}$

Answer and solution

Answer: (D) $\dfrac{1}{2}$

The minimum of

f(x) = x^2 - 4cx + 8c

is at

x = 2c

:

f_{\min} = 4c^2 - 8c^2 + 8c = -4c^2 + 8c

. The maximum of

g(x) = -x^2 + 3cx - 2c

is at

x = \dfrac{3c}{2}

:

g_{\max} = -\dfrac{9c^2}{4} + \dfrac{9c^2}{2} - 2c = \dfrac{9c^2}{4} - 2c

. We need

-4c^2 + 8c > \dfrac{9c^2}{4} - 2c

, i.e.

\dfrac{25c^2}{4} - 10c < 0

, i.e.

c(25c - 40) < 0

, so

0 < c < \dfrac{8}{5}

. Of the options

-2

,

-\dfrac12

,

2

and

\dfrac12

, only

\dfrac{1}{2}

lies between

0

and

1.6

(

2

is too large, and the negative values fail). The answer is

\dfrac{1}{2}

.

Q64TITALinear Equations

In a class, there were more than 10 boys and a certain number of girls. After 40% of the girls and 60% of the boys left the class, the remaining number of girls was 8 more than the remaining number of boys. Then, the minimum possible number of students initially in the class was

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 55

Let initial number of boys be $B$ and girls be $G$ .
Remaining girls = $0.6G$ . Remaining boys = $0.4B$ .
Given: $0.6G = 0.4B + 8 \Rightarrow 6G = 4B + 80 \Rightarrow 3G = 2B + 40$ .
Since 40% of girls and 60% of boys left, $G$ must be a multiple of 5, and $B$ must be a multiple of 5.
Let $B = 5x$ and $G = 5y$ .
$3(5y) = 2(5x) + 40 \Rightarrow 15y = 10x + 40 \Rightarrow 3y = 2x + 8 \Rightarrow y = \frac{2x + 8}{3}$ .
We need to minimize $B + G = 5x + 5y$ , so minimize $x + y$ .
Given $B > 10 \Rightarrow 5x > 10 \Rightarrow x > 2$ .
If $x = 3$ , $2(3)+8 = 14$ (not div by 3).
If $x = 4$ , $2(4)+8 = 16$ (not div by 3).
If $x = 5$ , $2(5)+8 = 18 \Rightarrow y = 6$ .
Minimum students = $5x + 5y = 5(5) + 5(6) = \mathbf{55}$ .

Q65MCQSequences & Series

In the set of consecutive odd numbers

\{1, 3, 5, \ldots, 57\}

, there is a number

k

such that the sum of all the elements less than

k

is equal to the sum of all the elements greater than

k

. Then,

k

equals

43
37
39
41

Answer and solution

Answer: (D) 41

The set

1, 3, 5, \dots, 57

has

\dfrac{57 - 1}{2} + 1 = 29

terms. The sum of the first

n

odd numbers is

n^2

, so the total is

29^2 = 841

. Let

k

be the

m

-th term, so

k = 2m - 1

and the terms before it add to

(m - 1)^2

. The terms after it must add to the same amount:

2(m - 1)^2 + (2m - 1) = 841

2m^2 - 2m + 1 = 841

, so

m^2 - m - 420 = 0

, i.e.

(m - 21)(m + 20) = 0

and

m = 21

. So

k = 2 \times 21 - 1 = 41

. Check: below 41 the sum is

20^2 = 400

; above it,

841 - 400 - 41 = 400

. Option C (39) is the 20th term: the sum below it is

19^2 = 361

, but the sum above it is

841 - 361 - 39 = 441

. Hence, option D (41).

Q66TITAProfit, Loss & Discount

A shopkeeper offers a discount of 22% on the marked price of each chair, and gives 13 chairs to a customer for the discounted price of 12 chairs to earn a profit of 26% on the transaction. If the cost price of each chair is Rs 100, then the marked price, in rupees, of each chair is

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 175

Let the marked price of one chair be $M$ . Cost price = 100.
Total CP for 13 chairs = $13 \times 100 = 1300$ .
Profit is 26%, so Total SP = $1300 \times 1.26 = 1638$ .
Customer buys 13 chairs for the discounted price of 12 chairs.
Discounted price per chair = $M(1 - 0.22) = 0.78M$ .
Total amount paid by customer = $12 \times 0.78M = 9.36M$ .
This must equal the total SP: $9.36M = 1638 \Rightarrow M = 1638 / 9.36 = \mathbf{175}$ .

Q67TITAPolygons & Circles

If the length of a side of a rhombus is 36 cm and the area of the rhombus is 396 sq. cm, then the absolute value of the difference between the lengths, in cm, of the diagonals of the rhombus is

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 60

Side of rhombus $s = 36$ cm, Area = 396 sq cm.
Area = $\frac{1}{2}d_1 d_2 = 396 \Rightarrow d_1 d_2 = 792$ .
Also, $d_1^2 + d_2^2 = 4s^2 = 4 \times (36)^2 = 4 \times 1296 = 5184$ .
We need $|d_1 - d_2|$ .
$(d_1 - d_2)^2 = d_1^2 + d_2^2 - 2d_1 d_2 = 5184 - 2(792) = 5184 - 1584 = 3600$ .
$|d_1 - d_2| = \sqrt{3600} = \mathbf{60}$ .

Q68MCQSequences & Series

For any natural number

k

, let

a_k = 3^k

. The smallest natural number

m

for which

\{ (a_1)^1 \times (a_2)^2 \times \dots \times (a_{20})^{20} \} < \{ a_{21} \times a_{22} \times \dots \times a_{20+m} \}

, is

57
56
59
58

Answer and solution

Answer: (D) 58

Since

a_k = 3^k

,

(a_k)^k = 3^{k^2}

. Left side:

3^{1^2 + 2^2 + \cdots + 20^2} = 3^{2870}

, because

\dfrac{20 \times 21 \times 41}{6} = 2870

. Right side:

3^{21 + 22 + \cdots + (20 + m)}

. This AP has

m

terms, so its sum is

\dfrac{m(41 + m)}{2}

. We need

\dfrac{m(m + 41)}{2} > 2870

, i.e.

m(m + 41) > 5740

. The positive root of

m^2 + 41m - 5740 = 0

is

\dfrac{-41 + \sqrt{24641}}{2} \approx 57.98

, so check the integers on either side:

m = 57

:

57 \times 98 = 5586 < 5740

, so the inequality fails.

m = 58

:

58 \times 99 = 5742 > 5740

, so it holds. Option A (57) falls just short: its exponent is

2793

, below

2870

. Hence, option D (58).

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