Practice Mock Test for SNAP - 11 - CAT MCQ

SP of first item = 0.75 x 200 = Rs. 150 and CP of this item = Rs. 80
CP and SP of the other item = Rs. x (as it is being sold at cost).

Total CP in the transaction = Rs. (80 + x)

Total SP in the transaction = Rs. (150 + x)

Since profit = 50%

150 + x= 1.5(80 + x)

150 + x= 120 + 1.5x

x = 60

Hence, option 2.

Let the distance between Ram and Shyam’s house be 10D km.

He walks 40% of the distance at 30% of his actual speed.

Therefore, he walks 4D km at 6 km/hr.
He walks 30% of the remaining distance at 50% of his speed. Therefore, he walks 1.8D km at 10 km/hr.
He walks the remaining distance at his actual speed.
Therefore, he walks 4.2D km at 20 km/hr.

Hence, option 2.

By observation,

The time required for Rs. 5600 to amount to Rs. 9800 will be the same as that required for Rs. 20,000 to amount to Rs. 35,000.
Hence, option 1.

Difference in Cl and SI in second year = Rs 32

Cl = SI for fist year = 800/2 = 400

This difference of Rs. 32 is the SI earned on Rs. 400 on one year (as the SI on first year is also added to principal for second year).

r= [(100 x 32)/(400 x 1)] = 8% Difference between Cl and SI for 3rd year = SI on Rs. 832 for one year

= (832 x 1 x 8)/100

Hence, option 2.

When three unbiased dice are thrown, there are 63 = 216 combinations possible.
The least possible sum is 3 and the highest possible sum is 18.

The multiples of 4 in this range are 4, 8, 12 and 16

Sum = 4:

Obtained for the combination (1,1, 2).

This combination can be made in 3!/2! = 3 ways

Sum = 8:

Obtained for the combinations(1, 1,6)- possible in 3!/2! = 3 ways (1,2, 5) - possible in 3! = 6 ways (1, 3, 4) - possible in 3! = 6 ways (2, 2, 4) - possible in 3!/2! = 3 ways (2, 3, 3) - possible in 3!/2! = 3 ways

Total number of ways = 3 + 6 + 6 + 3 + 3 = 21

Sum = 12:

Obtained for the combinations

(1, 5, 6) - possible in 3! = 6 ways

(2, 4, 6) - possible in 3! = 6 ways

(2, 5, 5) - possible in 3!/2! = 3 ways (3, 3, 6) - possible in 3!/2! = 3 ways

(3, 4, 5) - possible in 3! = 6 ways

(4, 4, 4) - possible in 1 way

Total number of ways = 6 + 6 + 3 + 3 + 6 + 1 =25

Sum = 16:

Obtained for the combinations

(4, 6, 6) - possible in 3!/2! = 3 ways

(5, 5, 6) - possible in 3!/2! = 3 ways Total number of ways = 3 + 3 = 6 Overall total unmber of ways = 3 + 21 + 25 + 6 = 55

Required probability = 55/216 = 0.254 Hence, option 1.

Let the speeds of P and Q be p and q respectively.
Given that P beats Q by 20 m. This means that the time taken by P to cover 100 m is same as that taken by Q to cover 80 m. Therefore, 4p = 5q.
Given that P and Q start simultaneously from A and B respectively, and they meet in 4 seconds.

Hence, option 1.

Given triangle is an isosceles triangle.
The third side = 4 cm or 10 cm

For a triangle, sum of any two sides is greater than the third side.

4 cm cannot be the third side, as 4 + 4 < 10

We can check that for 4 cm, 10 cm and 10 cm, sum of any two sides is greater than the third side.
The third side = 10 cm

Semiperimeter of the triangle = (4 + 10 + 10)/2 = 12 cm

Hence, option 2.

Let the number be x.

27x + 14x = 533 .

x = 533/41 = 13

Hence, option 2.

log₈108 = log₈(4 x 27)

log₈108 = log84 + log₈27

log₈4 = 2/3

log₈27 = (log₁₆27)/(log₁₆8)

log₈27 = (log₁₆27)/(3/4)

log₈27 = 4/3 (log₁₆27)

log₈27 = 4/3[(log₉27)/(log₉16)]

log₈27 = 4/3[(3/2)/(log₉16)]

log₈27 = 2/b

Iog₈108 = 2/3 + 2/b

 2(1/3 + 1/b)

= 2(b + 3)/3b

Hence, option 4.

Let there be x guests.

0.375x people are non-vegetarians. 0.625x are vegetarians.
Now, 0.375x are females

0.625x are males.
We have the following;

(m + n) + (0.625x - 300) + 300 = x

But we don't have any relationship between m and n.
Hence, we cannot uniquely determine the non-vegetarian female guest.
Hence, option 4.

2c = 9d = 9 x 13e

Hence, option 2.

Let the radius of an old wire be rand length of an old wire be h.
Radius of new wire = 2r and Length of new wire = H

Also, πr² h= π(2r)²H

H : h = 1 : 4 

Hence, option 1.

Where t is time, s is distance, v is velocity, Q is the amount of electricity used, n is the number of carriages and k₁,k₂ k care constants. Also, k = k₁,k₂

Hence, option 4.

Options 2 and 4 can be eliminated as the difference between the set of books is not always greater than 1.
Option 1: Total price = 5 x 1 7 + 8 x 1 9 + 1 7 x 2 3 = 628  620. Thus, this option is incorrect.
O ption 3: Total price = 5 x 1 7 + 1 0 x 1 9 + 1 5 x 2 3 = 620. Thus, this set is the correct number of books bought.
Hence, option 3.

Let the total amount be 7x.
A is supposed to get 4x and B is supposed to get 3x.
But actually A gets (1/7)^th part of 7x more i.e. x more.
Thus A gets 5x and B gets the remaining 2x.
Thus, the new ratio is 5 : 2.
Hence, option 2.

Neha keeps 5 bangles for herself, gives Meena 3 bangles and Kajol 2 bangles.
A group of p + q + r items can be divided into groups of p, q and r in (p + q + r)!/p! q! r! ways.
Number of ways in which she can divide the bangles = 10!/(5!3!2!) = 2520 ways.
Hence, option 3

Let A, B and C individually do a, b and c units of work per day. a + b = 2 c ... (i) Similarly, a + c = 3b ... (ii)

When all three work together, total work done per day = a + b + c
Using (i) and (ii), total work done per day (in units) = 3c = 4b

Since the three of them can finish the work in 19 days, let the total work be the LCM of 3, 4 and 19 i.e. 228 units

Total work done per day (in units) = 3c = 228/19 = 12

c = 4 i.e. C can do 4 units of work per day

Hence, if C were to work alone, C could complete the task in 228/4 = 57 days Hence, option 3.

Let the height of the tower be denoted by CD (as shown below with the angle of elevation and depression).

Y= 15m CD = 45 + 15 = 60 m Hence, option 3.

360 = 2³ x 3² x 5

Hence, option 4.

Rationalizing,

Hence, option 1.

The length of the circular track is 300 m.
Relative speed of Akbar and Amir = 10 + 15 = 25 m/s

Time after their start after which they meet = 300/25 = 12 s

Now, the distance between them is 300 m and they are running in opposite directions.
They will meet again after 12 s.
They will meet for the 30^th time after 12 * 30 = 360 s

Distance covered by Akbar when he meets Amir for the 30^th time = 15 x 360 = 5400 m = 5.4 km

Hence, option 2.

Let x people read both newspapers. (28 - x) + (35 - x) + x + 12 = 65

75 - x = 65

x = 10

Hence, option 1.

Solving (i) and (ii), x = 144 and y = 90 10% of (x + y) = 0.1 (144 + 90) = 23.4 Hence, option 1.

We know the formula,

Hence, option 2.

The highest power of 5 that can exactly divide 480! is the sum of the number of multiples of 5 in 480!, the number of multiples of 25 in 480! and the number of multiples of 125 in 480!.
The highest power of 5 that can exactly divide 480!

Hence, option 1.

In 60 days 40 men completed 2.5 km of the canal.
Let x number of men be required to complete 10 - 2.5 = 7.5 km of the canal in 180 - 60 = 120 days.

M₁ xD₁ x W₂ = M₂ x D₂ x W₁ 

60 x 40 x 7.5 = x x 120 x 2.5

x = 60 men

So the contractor requires 60 - 40 = 20 more men to complete the work on the scheduled time.
Hence, option 2.

x³ - 4x² + x + 6 = 0

This equation can be factorized in the form: ( x - 2)(x - 3)(x + 1)
So, the roots of this equation are 2, 3 and -1.
So a = 3, b = 2 and c = -1. ab + c= ( 3 x 2 ) + (-1 ) = 5

Hence, option 4.

Let the cost price of the article be Rs. 100.
Since the trader suffers a loss of 25% in the first transaction, his loss is Rs. 25 and consequently, the selling price is Rs. 75.
This selling price corresponds to a 40% discount i.e. the article is sold at 60% of the marked price.

In the second case, if the trader wants to make a profit of 10%, he should sell at Rs. 110.

Discount in this case = 125 - 110 = Rs. 15

Hence, option 2.

The above solution of alcohol has been successively diluted three times. Hence we use the formula for successive dilution.

Where v is the initial volume of pure alcohol, c is the concentration of the alcohol after replacements and x is the volume of alcohol replaced.

Amount of alcohol in the resultant solution = 0.512 * 100 = 51.2 litres Hence, option 3.

Let the heights of Prakash, Qaiser, Ronak, Santosh and Tarun be p, q, r, s and t respectively.

 It is given that,

 p + q + r - 154 x 3 = 462 ... (i)

p + q + r + s = 162 * 4 = 648 ... (ii)

s = 186 cm

 It is given that,

t = s - 10 = 176 cm

p + r + s + t = 580

p + r = 580 - (186 + 176) = 218 ... (iii)

Subtracting (iii) from (i), q = 244
The height of Qaiser = 244 cm

Hence, option 4.