Free CAT Quiz series 3 | CAT Online Test Series 3, CAT Free Mock Test

For any perfect square the index of the power of each of its prime factors is even. in the given number the power of 5 and 17 have odd indices, while 3 has an even index.
The least number = (5)(17) = 85.

Let the numbers be a and b. Then,
a + b = 12 and ab = 35.
(a + b)/ab = 12/35
= (1/b + 1/a) = 12/35
Sum of reciprocals of given numbers = 12/35.

Speed = 72 * 5/18 = 20 m/sec
Total distance covered = 110 + 132 = 242 m.
Required time = 242/20 = 12.1 sec.

624/6 = 104 kmph

2x² + 4x + x + 2 = 0
2x(x + 2) + 1(x + 2) = 0
(x + 2)(2x + 1) = 0 => x = -2, -1/2

The part of the tank filled by three pipes in one hour = 1/8
=> The part of the tank filled by two pipes in 1 hour = 2/3 * 1/8 = 1/12.
The tank can be filled in 12 hours.

C.P. of 1 kg = 420/7 = Rs. 6
S.P. of 1 kg = Rs. 6.50
Gain % = 0.50/6 * 100 = 25/3 = 8 1/3 %

5 > 4 and 1/2 > 1/4
5^1/2 > 4^1/4
4^1/4 cannot be the greatest. Hence 3^1/4 also cannot be the greatest.
3^7/10 = (3⁷)^1/10 = (2187)^1/10
5^1/2 = 5^5/10 = (5⁵)^1/10 = (3125)^1/10
6^1/5 = 6^2/10 = (6²)^1/10 = (36)^1/10
As (3125)^1/10 > (2187)^1/10 > (36)^1/10 , 5^1/2 is the greatest.

36 * 48 / 64 + 36/12 = 27 + 3 = 30

12 men take 32 days to complete the work.
One man will take (12 * 32) days to complete it.
Similarly one woman will take (16 * 36) days to complete it and one boy will take (48 * 16) days to complete it.
One man, one woman and one boy working together will complete = 1/(12 * 32) + 1/(16 * 36) + 1/(48 * 16)
= 1/(4 * 3 * 16 * 2) + 1/(16 * 4 * 9) + 1/(16 * 3 * 4 * 4)
= 1/64(1/6 + 1/9 + 1/12) = 1/64 * 13/36 of the work in a day.
They will take (64 * 36)/13 days to complete the work working together.

200*2 = 400
580
—–
180
1—-2
?—-180 = 90

Let the length of each train be x m.
(x + x) / (60 + 30)5/18 = (750 * 18) / (30 * 5) = 90 sec.

[50 – {20 + 60% of (26/13 + 24/12 + 21) – 2} + 8] = 625/x => [50 – {20 + 3/5 of (25) – 2} + 8] = 625/x
=> [50 – 33 + 8] = 625/x => x = 25

Let the radii of the smaller and the larger circles be s m and l m respectively.
2∏s = 264 and 2∏l = 352
s = 264/2∏ and l = 352/2∏
Difference between the areas = ∏l² – ∏s²
= ∏{176²/∏² – 132²/∏²}
= 176²/∏ – 132²/∏
= (176 – 132)(176 + 132)/∏
= (44)(308)/(22/7) = (2)(308)(7) = 4312 sq m

23 + 29 + 31 + 37 = 120

a³ = 1728 => a = 12
6a² = 6 * 12 * 12 = 864

Sum of the 35 numbers = 35 * 25 = 875
If each number is multiplied by 5, the sum also gets multiplied by 5 and the average also gets multiplied by 5.
Thus, the new average = 25 * 5 = 125.

760 —- 22
760 —- 20
x —– 19
x*19 = 760*20
x = 800
760
——-
40

A                              :                    B
(8000*4)+(4000*8) :    (9000*6)+(6000*6)
64000                     :    90000
32                           :    45

I. a² – 13a + 42 = 0
=>(a – 6)(a – 7) = 0 => a = 6, 7
II. b² – 15b + 56 = 0
=>(b – 7)(b – 8) = 0 => b = 7, 8
=>a ≤ b

3:1
2:1
——
6:1
1 —– 4000
7 —– ? => Rs.28,000

6 – [5/6 + {3 7/8 – 2 1/3} + 1 7/9] => 6 – [5/6 + {3 1/8 – 7/3} + 16/9]
=> 6 – [5/6 + {(279 – 168 + 128)/72}] => 6 – 5/6 + 239/72
=> 6 – (60 + 239)/72 => 6 – 46 – 11/72
=> 2 – 11/72 => 1 + 1 – 11/72 => 1 + 61/72 = 1 61/72

S.I. = (1200 * 10 * 1)/100 = Rs. 120
C.I. = [1200 * (1 + 5/100)² – 1200] = Rs. 123 Difference = (123 – 120) = Rs. 3.

As numerator is of the form a² – b² = (a + b) (a – b), so the given expression simplifying becomes
(0.9² – 0.8²)/(0.9 + 0.8) (0.9 + 0.8) (0.9 – 0.8)/0.9 + 0.8 = 0.1

A = 5000(51/50)³
= 5306.04
5000
———–
306.04

x * 3/8 = 141 => x= 376
376 * 32.08/100 = 120

8 * 84 = 672 – 7 = 665

T = 192/32 = 6 h
T = 6 – 3 = 3
Time Ratio = 6:3 = 2:1
Speed Ratio = 1:2

A:B = 2:3
B:C = 2:5
A:B:C = 4:6:15
6/25 * 75 = 18

a³ = 1331 => a = 11
b² = 121 => b = ± 11
a ≥ b

100 – 10 = 90
90——10
100——? => 11 1/9%

2: 5: 13
2x                  5x           13x
15                  7.5          2.5
30x + 37.5x + 32.5x = 100x/20x
Average = 5

P = 144(100/5)² => P = 3600

Rate of consumption of each man = 1050/(150 * 30) = 7/30 kg/day
Let us say 70 men take x days to consume 150 kg.
Quantity consumed by each item in x days = (7x/30) kg
Quantity consumed by 70 men in x days = (7/30 x)(70) kg
= (7/30 x) * (70) = 960
x = 60 days.

Amount of money Siddharth borrowed at S.I at 6% p.a. for two years = Rs.6,000
He lend the same amount for C.I at 6% p.a. for two years.
=> Siddharth’s income = C.I – S.I
= p[1 + r/ 100]ⁿ – p – pnr/100
= p{ [1 + r/ 100]² – 1 – nr/100
= 6,000{ [1 + 6/100]² – 1 – 12/100}
= 6,000 {(1.06)²– 1 – 0.12} = 6,000(1.1236 – 1 – 0.12)
= 6,000 (0.0036) = 6 * 3.6 = Rs.21.60

7/20 * 100 = 35

8820 —- 441
100 —- ? => 5%
x *105/100 * 105/100 = 8820
x*1.1025=8820
x=8820/1.1025 => 8000

x, 2x
(x-2):(2x-2) = 1:3
3x-6 = 2x-2
x = 4
x+2x = 3x
=> 3*4 = 12

(x*3*1)/100 + [(4000 – x)*5*1]/100 = 144
3x/100 + 200 – 5x/100 = 144
2x/100 = 56 → x = 2800

[50/x + 50/(x/2)] – 100/x = 2
x = 25

The ratio in which the four agencies will be paying the rents = 15 * 12 : 12 * 20 : 18 * 18 : 16 * 15
= 180 : 240 : 324 : 240 = 45 : 60 : 81 : 60
Let us consider the four amounts to be 45k, 60k, 81k and 60k respectively.
The total rent paid by the four agencies = 45k + 60k + 81k + 60k= 246k
It is given that A paid Rs. 1125
45k = 1125 => k = 25
246k = 246(25) = Rs. 6150
Thus the total rent paid by all the four agencies is Rs. 6150.

2000 + 300 + 40 + 5 = 2345

S = (100 + 300)/45 = 400/45 * 18/5 = 32

8/12 = 2/3 * 710 = 473.33
460.00
—————
13.33
1/3 uniform 13.33
1 ————— ? => Rs.40

Given that train P crosses a pole in 30 seconds and train Q crosses the same pole in one minute and 15 seconds.
Let the length of train P be L_P and that of train Q be L_Q
given that L_P = 3/4 L_Q
As the train P and Q crosses the pole in 30 seconds and 75 seconds respectively,
=> Speed of train P = V_P = L_P/30
Speed of train Q = V_Q = L_Q/75
L_P = 3/4 L_Q
=> V_P = 3/4 L_Q/(30) = L_Q/40
Ratio of their speeds = V_P : V_Q
= L_Q/40 : L_Q/75 => 1/40 : 1/75 = 15 : 8

Let C.P. of the article be Rs. x.
Then, required ratio = 104% of x / 106% of x
= 104/106 = 52/53 = 52:53

Here, five seniors out of 12 seniors can be selected in ¹²C₅ ways. Also, five juniors out of ten juniors can be selected ¹⁰C₅ ways. Hence the total number of different ways of selection = ¹²C₅ * ¹⁰C₅ = ¹²C₇ * ¹⁰C₅
= ¹²C₅ = ¹²C₇

Let the ten’s and unit’s digit be x and 8/x respectively.
Then,
(10x + 8/x) + 18 = 10 * 8/x + x
9x² + 18x – 72 = 0
x² + 2x – 8 = 0
(x + 4)(x – 2) = 0
x = 2
So, ten’s digit = 2 and unit’s digit = 4.
Hence, required number = 24.

Let the ten’s digit be x and unit’s digit be y.
Then, number 10x + y.
Number obtained by interchanging the digits = 10y + x.
(10x + y) + (10y + x) = 11(x + y)
which is divisible by 11.