MAT Online Test Series 2 | MAT Mock Test Series 2 | MAT Sample Papers |

P —- 2P —- 8 years
2 1/4 P —- 10 years

Let the train fare between the two places for one person be Rs.t
Bus fare between the two places for two persons Rs.4/3 t
=> 6/2 (4/3 t) + 8(t) = 1512
=> 12t = 1512 => t = 126.

Two balls can be picked from nine balls in ⁹C₂ ways.
We select one white ball and one red ball from five white balls and four red balls. This can be done ⁵C₁ . ⁴C₁ ways.
The required probability = (5 * 4)/⁹C₂ = 20/36 = 5/9

Clearly, (2272 – 875) = 1397, is exactly divisible by N.
Now. 1397 = 11 * 127
The required 3-digit number is 127, the sum of whose digit is 10.

8 = (1*20*R)/100
R = 40%

Sum of the present ages of husband, wife and child = (23 * 2 + 5 * 2) = 57 years.
Required average = 57/3 = 19 years.

Let B be turned off after x minutes. Then, part filled by (A + B) in x min + part filled by A in (30 – x) min = 1.
x(2/75 + 1/45) + (30- x) 2/75 = 1
11x + 180 – 6x = 225 => x = 9

P = 144(100/5)² => P = 3600

a² = 69696 => a = 264
4a = 1056
1056 * 15 = 15840

r₁: r₂ = 1: 3
Πr₁²: Πr₂²
r₁²: r₂² = 1: 9

Volume of the wire (in Cylindrical shape) is equal to the volume of the sphere.
π(16)² * h = (4/3)π (12)³ => h = 9 cm

25 * 2 * 0.75 = 20/100 * 10/100 * 7.5/100 * x
25 = 1/100 * x => x = 25000

X * (120/100)
—————- = 2/15
Y * (75/100)
X/Y = 1/12

Let the number of one rupee coins in the bag be x.
Number of 50 paise coins in the bag is 93 – x.
Total value of coins
[100x + 50(93 – x)]paise = 5600 paise
=> x = 74

7 * 8 : 9 * x = 8:9 => x= 7

Let A worked for x days.
x/21 + 4/28 = 1 => x/21 = 6/7 => x = 18
A worked for 18 days. So, A can complete the remaining work in 18 – 4 = 14 days.

(5m + 9w)10 = (6m + 12w)8
=> 50m + 90w = 48w + 96 w => 2m = 6w => 1m = 3w 5m + 9w = 5m + 3m = 8m
8 men can do the work in 10 days.
3m +3w = 3m + 1w = 4m
So, 4 men can do the work in (10 * 8)/4 = 20 days.

90 * 10/60 = 15 kmph

G^-1F⁺³I, Q^-2O⁺³R, L^-1K⁺³N, Y^-1X⁺³A and T^-1S⁺³V.
Except QOR, all other groups follows similar pattern.

Clearly, 252 = 2 * 2 * 3 * 3 * 7

Let ‘N’ is the smallest number which divided by 13 and 16 leaves respective remainders of 2 and 5.
Required number = (LCM of 13 and 16) – (common difference of divisors and remainders) = (208) – (11) = 197.

1/20 – 1/28 = 1/70
70 minutes

? = 17^14+16-8 = 17²²

HCF of 4956, 3894 = 354
(4956 – 3894)/354 = 3

Average = (76 + 65 + 82 + 67 + 85)/5 = 375/5 = 75.

Total fruits = 9
Since there must be at least two apples,
(⁵C₂ * ⁴C₁)/⁹C₃ + ⁵C₃/⁹C₃ = 25/42.

96% —- 12
144% —- ?
96/144 * 12 = 8

(3 * 4)/48 = (9 * 2)/ x
x = 72 kg

HCF of numerators = 1
LCM of denominators = 48
=> 1/48

Let the three parts A, B, C.
Then, A:B = 2:3 and B:C = 5:8
= (5 * 3/5) : (8 * 3/5) = 3:24/5
A:B:C = 2:3:24/5
= 10:15:24 => B = 98 * 15/49 = 30.

Let the sum be Rs. x. Then,
[x (1 + 4/100)² – x] = (676/625 x – x) = 51/625 x
S.I. = (x * 4 * 2)/100 = 2x/25
51x/625 – 2x/25 = 1 or x = 625.

Let the income and the expenditure of the person be Rs.5x and Rs.4x respectively.
Income, 5x = 18000 => x = 3600
Savings = Income – expenditure = 5x – 4x = x
So, savings = Rs.3600.

16 years ago, let T = x years and G = 8x years. After 8 years from now, T = (x + 16 + 8) years and G = (8x + 16 + 8) years.
8x + 24 = 3(x + 24) => 5x = 48
8 years ago, T/G = (x + 8)/(8x + 8) = (48/5 + 8) / (8 * 48/5 + 8) = 88/424 = 11/53

1/10 + 1/15 – 1/x = 1/18
x = 9

Let the length of the train be x m and its speed be y m/sec.
Then, x/y = 15 => y = x/15
(x + 100)/25 = x/15 => x = 150 m.

r = 3     πr² = 9
r = 5     πr² = 25
25 π – 16 π
100 —- ? => 64%

x + y = 80
x – 4y = 5
x = 65 y = 15

1000/71 = 14 6/71
1000 + 71 – 6 = 1065

37.5 % of 976
= 37.5/100 * 976 = 375/1000 * 976 = 3/8 * 976
= 3 * 122 = 366.

Let Rajan’s present age be x years.
Then, his age at the time of marriage = (x – 8) years.
x = 6/5 (x – 8)
5x = 6x – 48 => x = 48
Rajan’s sister’s age at the time of his marriage = (x – 8) – 10 = 30 years.
Rajan’s sister’s present age = (30 + 8) = 38 years.

20 = 4² + 4, 42 = 6² + 6, 58 = 7² + 9, 72 = 8² + 8 and 90 = 9² + 9.
20, 42, 72 and 90 can be expressed in n² + n form but not 58.

Work done by A and B in the first two hours, working alternately = First hour A + Second hour B = 1/4 + 1/12 = 1/3.
Total time required to complete the work = 2 * 3 = 6 days.

Let the price of each note book be Rs.x.
Let the number of note books which can be brought for Rs.300 each at a price of Rs.x be y.
Hence xy = 300
=> y = 300/x
(x + 5)(y – 10) = 300 => xy + 5y – 10x – 50 = xy
=>5(300/x) – 10x – 50 = 0 => -150 + x² + 5x = 0
multiplying both sides by -1/10x
=> x² + 15x – 10x – 150 = 0
=> x(x + 15) – 10(x + 15) = 0
=> x = 10 or -15
As x>0, x = 10.

Suppose, first pipe alone takes x hours to fill the tank. Then, second and third pipes will take (x – 5) and (x – 9) hours respectively to fill the tank.
1/x + 1/(x – 5) = 1/(x – 9)
(2x – 5)(x – 9) = x(x – 5)
x² – 18x + 45 = 0
(x- 15)(x – 3) = 0 => x = 15

41, 43, 47 and 53 are prime numbers, but not 57.

Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively.
Then, (2x + 4000)/(3x + 4000) = 40/57
6x = 68000 => 3x = 34000
Sumit’s present salary = (3x + 4000) = 34000 + 4000 = Rs. 38,000.

Given that, the question paper consists of five problems. For each problem, one or two or three or none of the choices can be attempted.
Hence, the required number of ways = 4⁵ – 1.
= 2¹⁰ – 1 = 1024 – 1 = 1023

7/20 * 100 = 35

M = 4.5
S = x
DS = 4.5 + x
US = 4.5 + x
4.5 + x = (4.5 – x)2
4.5 + x = 9 -2x
3x = 4.5
x = 1.5

Let C.P. = Rs. x.
Then, S.P. = Rs. (12% of x) = Rs. 6x/5
New S.P. = 2 * 6x/5 = Rs. 12x/5
Profit = 12x/5 – x = Rs. 7x/5
Profit = 7x/5 * 1/x * 100 = 140%.