Practice Mock Test for SNAP - 9 - CAT MCQ

x! = y! + y x y! x!

= (y + 1) x y| x!

= (y + 1)!

Thus, x and y are consecutive integers.
However, one cannot say whether x is even or odd.
Thus, the question cannot be answered using statement A alone.

Using Statement B alone:

• No information about x is given.
• Thus, the question cannot be answered using statement B alone. Using both the statements together: Since y is a multiple of 6, it is an even number.
• x must be odd.

Thus, the question can be answered using both the statements together but not by using either statement alone.
Hence, option 3.

Each of A, B and C won at least one match out of 3 matches.
Hence they should have won exactly one match.
After round 3 each of them had Rs. 27 with them.
Let C loses round 3, thus he would have paid the other two players twice the amount which they already have i.e., he would have tripled the money of A and B.

Thus, before round 3 or at the end of round 2, A and B would have got Rs. 9 with them while C would have got Rs. 63 with him.
Let B loses Round 2, thus he would have tripled the money of A and C.
Thus, before round 2 or at the end of round 1, A and C would have got Rs. 3 and Rs. 21 respectively while B would have got Rs. 57 with him.
Let A loses Round 1, thus he would have tripled the money of B and C.
Thus, before round 1 or at the beginning, B and C would have got Rs. 19 and Rs. 7 respectively while A would have got Rs. 55 with him.
Thus A would have got Rs. 55 at the beginning of the game.

Hence, option 1.

Given:

• Average milk consumption on weekdays (total 22 days) is 3 litres per day.
• Total milk consumed on weekdays = 22 x 3 = 66 litres
• Total milk delivered in 1month = 98 litres.

So total milk delivered (and consumed) on weekends = 98 - 66 = 32 litres

Number of weekend days in April:

= Total days - Number of weekdays
= 30 - 22 = 8

Average milk consumption on weekends:

= 32/8
= 4 litres.

Hence, option 2.

15^7k-3 > 1

► 15^7k-3 > 15⁰
► 7k-3 > 0
► k > 3/7

Hence, option 2.

log₆ 169 x log₁₃ 6 x log₇ 64 x log₄ 49

= (log 13² / log 6) x (log 6 / log 13) x (log 4³ / log 7) x (log 7² / log 4)
= 2 x 1 x 3 x 2
= 12

Hence, option 3.

1.4x - x = x * p * n / 100
p * n = 40

Required simple interest:

= 14200 x 40 / 100
= 5680

Hence, option 3.

Let the average weight of the 59 students be A.
Therefore, the total weight of the 59 of them will be 59 A.
The questions states that when the weight of this student who left is added, the total weight of the class
= 59A + 45
When this student is also included, the average weight decreases by 0.2 kgs.
59A + 45 / 60 = A - 0.2
⇒ 59A + 45 = 60A - 12
⇒ 45 + 12 = 60A - 59A
⇒ A = 57

Consider annual compounding.
Interest for the first year is the same, irrespective of whether it is simple interest or compound interest.
Hence, the difference between the two interest amounts is due to the extra compound interest earned in the second year.
The C.l. earned in the first year is added to the original principal and this sum becomes the principal for the second year.
Hence, the situation is equivalent to placing (original principal + C.l. of first year (say x)) at 8% simple interest for one year.
Hence, effectively, Rs.30.72 is the simple interest earned by placing Rs. x at 8% S.l. for a year 30.72 = x(0.08) x = 384
This is also the S.l. earned by keeping the original principal at 8% for a year
Original principal = 384/0.08 = 4800 Hence, at half-yearly compounding, amount due at the end of 2 years = 4800[1 + (4/100)]⁴ = 5615.32

C.l. = 5615.32 - 4800 = Rs. 815.32

S.l. for Rs.4,800 at 8% for 2 years = (4800 x 8 x 2)/100 = Rs. 768

Required difference = 815.32 - 768 = Rs. 47.32

Hence, option 3.

The given expression is:
(5√-1024)³ + (4√2401)³ + (3√-27)³ 
Now, (-4)⁵ = -1024, (-3)³ = -27 and (±7)⁴ = 2401
Hence, the given expression is of the form (-4)³ + (±7)³ + (-3)³ = - 64 ± 343 - 27 = 252 or -434 Among these two values, only 252 is in the options.
Hence, option 2.

Cost of wooden flooring = Rs. 750 Cost of Italian marble flooring = 60% of 750 = Rs. 450

Area of the living room = 28 x 25 = 700 sq m

Area of the bedroom = 20 x 18 = 360 sq m

Area o f the kitchen = 12 x 10 = 120 sqm

Total cost of the wooden flooring = 700 x 750 = Rs. 5,25,000 Total cost of Italian marble flooring = (360 + 120) x 450 = Rs. 2,16,000
Required difference = 525000 - 216000 = Rs. 3,09,000

Hence, option 2.

Area of the bedroom with ceiling when the height of the bedroom is 12 m = 2(20 x 18 + 20 x 12 + 18 x 12)

= 2(360 + 240 + 216)
= 1632 sq m

Out of this, 360 sq.m on the floor has to have flooring.
Area to be painted = 1632 - 360 = 1272 sq.m

Total cost of painting = 1272 x 315 = Rs. 4,00,680

Total cost of flooring the bedroom = 360 * 450 = Rs. 1,62,000
Total cost of renovating the bedroom = 400680 + 162000 = Rs. 562680 Hence, option 4.

Total area of the house = 1300 sq m Total area of the house oringinally to be renovated = 700 + 360 + 120 = 1180 sq m.
Remaining area of the house = 1300 - 1180 = 120 sq m

Total cost of flooring one-third of the required area with Italian marble = 40 x 450 = Rs. 18,000

Total cost of flooring the remaining area with wood = 80 * 750 = Rs. 60,000

Total increase in the original cost of flooring = 18000 + 60000 = Rs. 78,000 Hence, option 4.

The first 28 multiples of 28 are 28, 28 * 2, 28 x 3 , 28 * 28.
The arithmetic mean of these 28 numbers would be the sum of all the numbers divided by 28.
AM = 28 x (28 x 29 / 2) / 28 = 406
Hence, option 2.
Alternatively,
The arithmetic mean of the first 28 multiples will be equal to the arithmetic mean of the 14^th and 15^th multiple.
AM = 392 + 420 / 2 = 406
Hence, option 2.

Let the 1st term of the arithmetic progression be a and the common difference be d.

The 5^th term will be (a + 4d) and the 18^th term will be (a + 17d).
Also, the 7^th term will be (a + 6d), the 12^th term will be (a + 11d) and the 15^th term will be (a +14d).
Now, from the given condition, we get (a + 4 d) + (a + 17d) = (a + d) + (a + 11d + (a + 14d)

2a + 21d = 3a + 31d

 a + 10d = 0 (a + 10d) is the 11^th term of the series.

Let x₁, x₂, x₃, x₄ and x₅ be the number of prizes given to the 5 boys respectively.
We have x₁ + x₂ + x₃ + x₄ + x₅ = 101 Since each boy received odd number of prizes, let
x₁ = 2y₁ - 1, x₂ = 2y₂ - 1, x₃ = 2y₃ - 1, x₄ = 2y₄ - 1, x₅ = 2y₅ - 1
where y_1, y_2, y₃, y₄ and ys are positive integers
So we have (2y₁ - 1) + (2y₂ - 1) (2y₃ - 1) (2y₄ - 1) (2y₅ - 1) = 101
i.e y₁ + y ₂ + y₃ + y₄, + y₅ = 53
Now imagine that these 53 prizes are placed one beside the other. So we have 52 spaces in between these prizes. To divide them into 5 groups, we need to select 4 spaces from the available 52 spaces which can be done in 52_C4 ways. Hence, option 1.

The big pile of apples contains the same number of large apples of half a rupee each (from Savithri), as smaller apples of one third a rupee each (from Gayathri).
∴ The average price  
= (1 / 2 + 1 / 3) / 2 = 5 / 12 rupees

The apples should be sold at Rs. (5/12) each.
But the apples were sold for Rs. (2/5) each (5 apples for Rs. 2).

Each apple is sold at Rs. (1/60) less.
The total shortage is Rs. 7,

After combining there were 7 x 60 = 420 apples.

The amount obtained  from these 420 apples = 2/5 x 420 = 168 rupees 
They divided the money equally amongst themselves, each of them got Rs. 84.
If Savithri would have sold the apples herself, she would have received Rs. 105 for 210 apples.
Savithri loses Rs. 21 in this deal.
Hence, option 2.

Given series: 13, 10, 5, 13, 37, 98, 220. Consider the first level difference:

10 - 13 = - 3 = 0³ - 3

5 - 10 = -5 (It should be -2 = 1³ - 3)

13 - 5 = 8 (It should be 5 = 2³ - 3)

37 - 13 = 24 = 3³ - 3
98 - 37 = 61 = 4³ - 3
220 - 9 8 = 122 = 5³ - 3
Here, it can be seen that for the 3^rd term i.e. 5 the pattern differs from what it needs to be. (The 3^rd term should be 8).
The term 5 is the odd one out.

Hence, option 3.

Let the initial property of Naveen be Rs. x.
As per given conditions,
x - 35x / 100 - 30x / 100 = 30
∴ 35x / 100 = 30
x = Rs. 85.7 lakhs
Hence, option 3.

The largest five digit number is 99999.
Now, by actual division, 99999 divided by 86 leaves remainder 67.
The largest five digit number, divisible by 86, is 99999 - 67 = 99932
Hence, option 1.

Alternatively, We can easily see that 86860 is divisible by 86. 8600, 860 and 86 are all divisible by 86 too.
So, 86860 + 8600 = 95460 is divisible by 86. 95460 + 8600 > 99999
95460 + (5 x 860) = 99760 is divisible by 86. 99760 + 2 x 86 = 99932 is divisible by 86.

99932 + 86 > 99999

99932 is the largest number divisible by 86.

Hence, option 1.

Let the distance between Thane and Kalyan be x km.
Time taken by Sajan to cover x km at 24 kmph = (x / 24) hours.
Time taken by Samir to cover x km at 32 kmph = (x / 32) hours. 
It is given that Samir reaches 35 minutes = (7 / 12) hours before Sajan,
(x / 24) - (x / 32) = (7 / 12)
4x - 3x = 56
x = 56 km
Hence, option 1.

Let the initial amount of wine in the cask be W, then
135 = W(1 - 25 / 100)³
135 = W(3 / 4)³
W = 135 x 64 / 27
W= 320 litres
The initial amount of wine in the cask is 320 litre.
Hence, option 4.

Total water available for 140 members for 90 days = 140 x 90 units
Total available consumed by 140 members in 50 days = 140 x 50 units
Water available after 50 days = 140(90 - 50) = 140 x 40 units

Number of members after 50 days = 160
∴ Number of days that the water will sufficient = 5600/160 = 35days

Hence, option 4.

Let the first term and the common difference of this arithmetic progression be a and d respectively.
Sum of first 6 terms = 6 / 2 x [2a +(6 - 1)d] = 3[2a + 5d] ...(i)
Sum of first 6 terms = 12 / 2 x [2a + (12 - 1)d] = 6[2a + 11d]...(ii)
Divide (i) by (ii) to get
3[2a + 5d] / 6[2a + 11d] = 37 / 134
Hence we get a : d = 6 : 5
Applying componendo and dividendo we get,
a + d / a - d = 6 + 5 / 6 - 5 = 11
Hence, option 4.

The selling price of the fabric is Rs. 405 which corresponds to 50% profit on the cost price. 405 = 1.5 x CP. CP = 405/1.5 = Rs. 270. Now, the ratio of the weight of cotton to that of wool in the fabric is 3 : 7

So, the fabric is 30% cotton and 70% wool.
Let the per kg cost price of cotton and wool be Rs. 2x and Rs. 3x respectively.
(0.3 x 2x) + (0.7 x 3x) = 270

2.7x = 270

x = 100.

Price of the cotton in the fabric = 2 * 100 * 0.3 = Rs. 60

Hence, option 3.

Since the journey covered in both cases is the same, the distance is constant. So, the speed and time are inversely proportional to each other.
Let the original speed of Ramesh be s km/min and the original time taken be t minutes.
Since he travels 33.33% faster, his speed today is 1.33s i.e. (4/3)s km/min and the new time taken is (t - 45) minutes.

st = (4s/3) x (t - 45)

:. 3t = 4t - 180

t = 180
So, time taken at new speed = t - 45 = 180 - 45 = 135 minutes i.e. 2.25 hours.
Hence, option 2.

Let the original cost price and selling price be x and y.
Intial Profit Percentage = (y - x / x) x 100
New Profit Percentage = (0.84y - 1.2x / 1.2x) x 100 = 60
(y - x / x) x 100 - (0.84y - 1.2x / 1.2x) x 100 = 60
⇒ y / x (1 - 84 / 120) = 3 / 5
⇒ y / x x 36 / 120 = 3 / 5 ⇒ y / x = 2 / 1
Initial Profit Percentage = y - x / x x 100 = 100%
Hence, option 4.

2 adults have to be seated at the 2 ends. This can be done in ³P₂ ways = 6 ways

The remaining adult and all children occupy the remaining 6 seats in such a way that the 2 girls are together.
Considering the 2 girls as a group, we can place the adult, 3 boys and 1 group of girls in 5 places in 5! Ways = 120 ways

The girls can be further placed in 2! ways within the group = 2 ways
Required number of ways = 6 * 120 x 2 = 1440 ways

Hence, option 2.

Let the marks obtained by B be 100.
Therefore, marks obtained by A will be 100 - 20% of 100 = 80. Marks obtained by B were 40% less than that obtained by C.
Therefore, marks obtained by C
= 100 / 1 - 0.4 = 100 / 0.6 = 500 / 3
Required percentage = 80 / 500 / 3 x 100 = 48%
Hence, option 2.

2^{(m - n)} = 16 = 2⁴ ⇒ m - n = 4
3^{(m + n)} = 81 = 3⁴ ⇒ m + n = 4
Solving (i) and (ii), we get m = 4 and n = 0.
Hence, option 2.

The fox covers (17/6) x 3 = 8.5 m, when the deer covers 1 x 8 = 8 m
i.e., when the fox takes 3 leaps, he covers 0.5 m of the gap.

To cover 2.5 m, it will have to make 5 x 3 = 15 leaps.
Hence, option 3.