NMAT Quantitative Skills MCQ Quiz - 1 - CAT MCQ

Represent Quant-lengthy-easy as Q-L-E, Verbal-non-lengthy-difficult as V-NL-D and so on.
Let Q-L-E = 5x, Q-NL-E = 2z, Q-L-D = K and V-L-E = y

V-L-D = 3x (from 2), V-NL-D = z - 10 (from 1), V-NL-E = 3K (from 6) and Q-NL-D = 3y - 5 (from 3)

Now, the questions in Quants and Verbal can be classified as easy/difficult or lengthy/non-lengthy as shown below.

The number of non-lengthy questions in Quants is 30 more than the number of difficult questions in Verbal.

2z + 3y - 5 = (3x + z — 10) + 30

3y + z - 3x = 15 ...(i)

The number of lengthy questions in Verbal section is 35 less than the number of easy questions in Quant.

y + 3x = (5x + 2z) - 35

2x - y + 2z = 35 ...(ii)

From (i) and (ii)

7y - 8x = 15

(x,y) = (6, 9) or (13, 17) or (20, 25) etc.

Corresponding value for z = 16 or 13 or 10 etc..
Since each sub-section has at least 5 qestions, the number of non-lengthy, difficult question in the verbal section
(z - 10) > 5.
z > 15
z = 16
Thus, (x,y ,z) = (6, 9, 16) Also,the number of question in Quants is less than 100 while that in verbal is not less than 75.
Number o f questions in Quants = 5x + 2z + (3y - 5 ) + K = M + K < 100 

Consider the solution to first question. Number of questions in each subsection given in the options is: Option 1: V-L-E =y = 9

Option 2: Q-NL-D = 3y - 5 = 3(9) - 5 = 22

Option 3: V-NL-E = 3K= 3(14) or 3(15) = 42 or 45

Option 4: Q-L-D = K= 14 or 15
Option 5: V-NL-D = z - 10 = 16 - 10 = 6 Hence, option 5.

Consider the solution to the first question.

Number of easy questions = 5x + 2z + y + 3K

(x,y,z) = (6, 9, 16)

Number of easy questions = 71 + 3K K= 14 or 15

Number of easy questions =113 or 116 Hence, option 1.

Consider the solution to the first question.
Total questions = 5x + 2z + K + 3 y ~ 5 + y + 3K +3 x + z - 10

= 8x + 4y + 3z + 4K - 15

(x,y,z) = (6, 9, 16)

Total questions = 8(6) + 4(9) + 3(16) + 4AT— 15 = 117 + 4K

K= 14 or 15

Maximum number of questions =117 +(4 x 15) =177

The score is maximized if the topper answers each question correctly.
Maximum score = 177 x 4 = 708 Hence, option 5.

Since this is always true, x + 3 > 0 i.e. x > -3

Since this is never true, x + 3 is not less than 0.
Except for x = -4, all the other values satisfy x > -3.
Hence, option 3.

Hence, option 3.

The largest number cannot be 1 or 2 as is the largest of three consecutive natural numbers.
Now, consider the options.
If the largest number is 3, the numbers are 1,2 and 3. Their product is 6. It exceeds the cube of 1 by 5, is less than the cube of 2 by 2 and is less than the square of 3 by 3.
Thus 3 satisfies all the conditions.
Hence, option 1.

Total volume of the block of ice-cream = (Volume of the hemispherical scoop) x 2 x (no. of people served)

Let x people get the ice cream.

Hence, option 2.

Rohan’s 25^th birthday will be on 29^th March 2015.
Now, the first 2000 completed years comprise 5 sets of 400 years each. Since there are 0 odd days in 400 years, there are 0 odd days in the first 2000 years as well.
The 14 completed years from 2000 to 2014 have 3 leap years (2004, 2008 and 2012) and 11 non-leap years.
Since a non-leap year has 1 odd day and a leap year has 2 odd days,

Total odd days in these 14 years = 11(1) + 3(2) = 17 = 7(2) + 3 i.e. 3 odd days.

Total number of days in 2015 (upto 29^th March) = 31 + 28 + 29 = 88 days (as 2015 is not a leap year).
Number of odd days in this period = 7(12) + 4 i.e. 4 odd days.
So, total odd days till Rohan’s birthday = 3 + 4 = 7 i.e. 0 odd days.
Hence, Rohan’s 25^th birthday will fall on a Sunday.
Hence, option 3.

If we try to solve this problem by using distance formula, it will require a lot of calculations.
Let us approach this problem in a different way.

‘C’ is the mirror image of B(9, 14)

Now the path taken by the ant is APB, where P(x, 0).
We know that PE = PC

We can safely say that path taken by the ant is APC.
And we need to find minimum length of APC.
APC will be minimum if A, P, C lie on a same straight line,

i.e. AC = AP + PC

Equation of line AC is:  

Hence, option 4.

 Weight of Zebesco that does not come from protein in tissue = 21 -1.638 = 19.362 stone

Hence, option 3.

Weight contributed by protein in skin = 0.08 x 0.15 x 52 = 0.624 kg

The difference between this value and the weight contributed by fat can be maximized when fat contributes the least to ‘other nutrients’.

 Weight contributed by fat = (1/10) x 0.15 x 52 = 0.78 kg

Maximum difference = 0.78 - 0.624 = 0.156 kg = (0.156/6.3) stone ~ 0.02 stone

Hence, option 1.

The weight due to protein in muscles can be found. However, it is not known if water or other nutrients also contribute to muscle weight, and if so, in what proportion.
Hence, the body weight due to muscles cannot be found.
Hence, option 5.

1 stone > 1 kg > 1 pound

Also 1 stone =14 pounds

Convert all the weights given into a common unit, say stone.
Option 1: 52% of 15% of 13.5 stone Option 2: 40% of 15% of some amount less than 10 stone (because 10 stone  kg)

Option 3: 70% of 10.3 stone

Option 4: 30% o f some amount less than 9 stone (because 9 stone = 9 x 1 4 pounds =126 pounds)

Option 5: 23% of 15% of some amount less than 8 stone (as explained in option 4).
By observation itself, options 2, 4 and 5 will be less than option 3.
Also, option 3 is nearly 7 stone while option 1 is not even 2 stone.
Hence, the weight in option 3 is the greatest.
Hence, option 3.

Minimum monthly wheat consumption in 2011 = 15100 tons

Maximum monthly wheat consumption in 2011 = 17400 tons

Difference between the maximum and minimum wheat consumption for 2011 = 2300 tons

Minimum monthly rice consumption in 2012 = 18200 tons

Maximum monthly rice consumption in 2012 = 25500 tons

Difference between the maximum and minimum rice consumption in 2012 = 7300 tons

Hence, the required ratio = 23 : 73  1 : 3

Hence, option 3.

Consider the monthly cereal consumption ( in 000 tons) in 2012 for only the five months mentioned in the answer options.

 Aug = 18.1 + 13.9 + 25.5 = 57.5

Sep = 17.4 + 13.8 + 23.4 = 54.6

Oct = 17.6 + 14.3 + 24.5 = 56.4

Nov = 17.8 + 15.3 + 23.6 = 56.7

Dec = 16.1 + 15.2 + 25.1 = 56.4

Hence August has the maximum cereal consumption in 2012.
Hence, option 5.

A negative percentage change in the monthly consumption basically implies that the consumption in a particular month is less than the consumption in the previous month.
The number of times there is a negative percentage change for each option is:
Wheat in 2011 : 4 times - Mar, June, Sept, and Dec

Jwari in 2011 : 2 times - April and Oct Rice in 2012 : 5 times - Mar, April, Jun, Sept and Nov

Jwari in 2012 : 5 times - June, Jul, Aug, Sept and Dec

Thus, the rice consumption in 2012 as well as the jwari consumption in 2012 both show the most instances of a negative percentage change.
Hence, option 5.

Here, observe that the absolute change in the rice consumption in 2011 was less than 2000 tons for every consecutive month apart from December.
Between November and December, the rice consumption increased by 2400 tons.

It can be verified that the percentage change in all other months was less than this value.
Hence, option 1.

The figure is as shown below.

The cow is grazing along the circle with centre as P while the field is a quarter with centre at O. Let the smaller circle touch the bigger circle along its circumference at Z.

Now, let the radius of bigger circle be R and that of smaller circle be r.
Now, CP = DP = r

Now, if we extend line OP to meet the circle at Z, then

Note that we need to paint only curved surface areas of hemisphere and cylinder.
Curved Surface Area of the hemisphere = 2πr²

Curved Surface Area of the right circular cylinder = 2πrh

Total surface area that is to be painted = (308 + 440) cm² = 748 cm²

Cost of painting the toy = 748 x 14 = Rs. 10,472

Hence, option 3.

Aran takes x + 18 days and Ram takes x + 8 days to complete the work

Working together, they finish the work in x days.

Solving this equation, x = 12

Hence, working together, Arun and Ram complete the work in 12 days.
Individually, Aran takes 30 days and Ram takes 20 days.

Hence, option 2.  

Since AD is the median to BC, BD - CD - BC/2 - 12/2 - 6 cm

According to the Apollonius’ Theorem, AB² + AC² = 2(AD² + BD²)
12² + 8² = 2(AD² + 6²)
144 + 64 = 2 AD² + 72

Hence, option 4.

Each student attempted all the questions and no two students had the same answer key.
Since each question had 5 options, each question could have been answered in 5 different ways.
Since there were 5 such questions, total possible answer keys = 5⁵ = 3125, which exceeds the number of students (i.e. 2500).
Hence, only 2500 of the possible 3125 answer keys will be used.
Minimum number of students getting a positive score = Total number of students - Maximum number of students who can get a zero or a negative score
Since every correct and incorrect answer gives +4 and -1 marks respectively, a student should answer 1 question correctly and 4 questions incorrectly to get a score of 0.
The question that is answered correctly can be selected in 5 different ways.
For any question that is answered incorrectly, the student can select one out of the four incorrect options in 4 different ways.
Since there are four questions answered incorrectly, total ways to get a score of0 = 5 x4⁴ = 5 x 256= 1280 Hence, maximum number of students who can get a score of 0 = 1280

To get a negative score, a student should answer all the questions incorrectly.

As explained above, one question can be incorrectly answered in 4 ways.
Since all 5 questions are incorrectly answered, total ways to get a negative score = 4⁵ = 1024. This is the maximum number of students who can get a negative score.
The minimum number of students getting a net positive score = 2500 - (1280+ 1024) = 196

Hence, option 2.

The total number of chocolates has to lie between 85 and 95 and has to be divisible by 8+1=9 as well as 2+1=3. The only number satisfying these is 90.
Result I: Ram 80, Lakhan 10
Result II: Ram 30, Lakhan 60
Number of chocolates with Ram = (80+30)/2 = 55

Option (4).    1200 is the correct answer. 

Explanation:- 

We need to find the no. Which is of the form aba^2b^2cd ; where a,b,c and d can take any value from 0 to 9.

a and b cannot exceed 3 as the square of the two numbers then would exceed 9, which is not possible.

A can have 3 values ( 1,2 or 3) and b can have 4 values ( 0,1,2 or 3) c and d can have 10 values each. 

Thus the total count of numbers that satisfy the given conditions is :- 3*4*10*10 

=> 1200 es each.

Thus total count of numbers that satisfy the given conditions is 3 x 4 x 10 = 1200

Revenue generated from 220 tonnes of saw timber in 2014 = 220 x 19000 = Rs. 41.8 lakhs

Quantity of plywood sold (in tonnes) in 2011 to generate the same revenue = 4180000/5000 = 836
Since 800 kg of plywood corresponds to 1 cubic metre, (836 x 1000) kg of plywood corresponds to (836 x 1000)/800 cubic metre i.e. 1045 cubic metre

Hence, option 2.

For plywood and saw timber, 1 cubic metre = 800 kg = 0.8 tonne

If the selling price of 1 tonne of plywood or saw timber is Rs. jc, the selling price of 0.8 tonnes is Rs. (0.8x) For any year, total selling price per tonne = logs + 0.8(plywood) + 0.8(saw timber) = logs + 0.8(plywood + saw timber)

Since one of the options is ‘none of the above’, calculate the total selling price for each year.
By observation, all three products have their least selling price in 2008. Hence, the overall least selling price is also in 2008.
Hence, option 5.

The selling price of logs is given per cubic metre but the conversion from cubic metre to tonnes is not given for logs.
Hence, the selling price per tonne cannot be found for logs.
Hence, the required ratio cannot be found.
Hence, option 5.

Consider the solution to the earlier questions.
If selling price of plywood or saw timber = Rs. x per tonne, then selling price = Rs. (0.8x) per cubic metre.

In 2012, selling price per cubic metre (in 000s) for all three products is:

logs =18, plywood = 0.8 x 4 = 3.2 and saw timber = 0.8 x 13 = 10.4

Now, let 100 cubic metres be sold in all.
Total selling price = (20 x 3.2) + (30 x 10.4) + (50 x 18) = 64 + 312 + 900 = 1276
Average selling price (in Rs. 000s) per cubic metre = 1276/100 = 12.76  13

Hence, option 4.

Savings = Income - (Utility + Rent + Grocery + Misc)

The savings for each month are as shown in the table above.

Thus, the saving are highest in May.

Hence, option 3.