NMAT Quantitative Skills MCQ Quiz - 4 - CAT MCQ

Using statement A alone: Only the angle of inclination is known. Since the height of each floor is unknown, the height of the building cannot be found. So, the length of the ladder cannot be found. 

Thus, the question cannot be answered using statement A alone.
Using statement B alone: Since the ladder is inclined at 45° with the ground, the angle of elevation is 45°.
Hence, the height of the fifth floor of the building is equal to the distance between the base of the building and ladder.
Thus, the triangle so formed is a right isosceles triangle (with sides equal to 47.89 m) and the length of the ladder is its hypotenuse.
So, the length of the ladder can be found.
Thus, the question can be answered using statement B alone.
Thus, the question can be answered using statement B alone, but cannot be answered using statement A alone.

Hence, option 2.

Using Statement I alone: 12x - 4y = 48 Multiply the above equation by 2.5 to get: 30x - 10y = 120 Thus for the two equations, we get:

Thus, infinite solutions exist for the given equation and a unique value of x and j cannot be found.
Thus, the question cannot be answered using statement I alone.
Using Statement II alone:32^x = 45^y
This equation is satisfied only for x =y = 0 Thus, the question can be answered using statement II alone.
Thus, the question can be answered using statement II alone but not by using statement I alone. Hence, option 2

Using statement I alone: SP = 2CP  

Both the statements independently answer the question.
Hence, option 3.

Thus, statement A alone is sufficient to answer the question.
Using statement B alone:

Thus, statement B alone is not sufficient to answer the question. Hence, option 1.

Using Statement A alone: 

Area of triangle = abc\4r

where a, b, c are the sides of the triangle and R is the circumradius Since the length of sides is not known, the area cannot be found. Thus, the question cannot be answered using statement A alone.
Using Statement B alone: Since the triangle has the maximum possible area, it has to be an equilateral triangle Hence, each side of the triangle = 36/3 = 12 cm Once the length of each side of a triangle, its semi-perimeter and area can be found.
Thus, the question can be answered using statement B alone.
Thus, the question can be answered using statement B alone but not by using statement A alone.
Hence, option 2.

Using statement A alone: Since 6th April 2064 will be a Friday, 7th, 14th, 21st and 28th April 2064 will be Saturdays.
So, the last Saturday of April 2064 will be on the 28th. 

Thus, the question can be answered using statement A alone.
Using statement B alone: Since the last day of April 2064 will be a Thursday, 30th April will be a Thursday.
Thus, 25th April will be the last Saturday of the month.
Thus, the question can be answered using statement B alone.
Thus, the question can be answered using either statement alone.
Hence, option 3.

Let the total work be the LCM of 17 and 10 i.e. 170 units.
Let A do a units of work per day.
Since B works at half or one-fourth of his normal efficiency, let B do 4b units of work per day.
In the first case, work is completed in 17 days. 17(2a + 2b) = 170. 2a + 2b= 10.  a + b = 5 ... (i) In the second case, work is completed in 10 days. 10(5a + b) = 170  5a + b = 17 ... (ii)

Solving (i) and (ii), a = 3 and b = 2. Hence, amount of work done by B per day = 4b = 8 units

Hence, option 1.

Basically we need to find out the count of 4-digit number of the form: [m x L.C.M.(7, 9) + 3] i.e. of the form 63m + 3, where m is a natural number.
The total numbers below 10000 and of the form 63m + 3 are 158 and the largest such number is (63 x 158 + 3) = 9957. Similarly the total numbers below 1000 and of the form 63m + 3 are 15 and the largest such number is (63 x 15 + 3) = 948 Thus the number of 4-digit numbers which when divided by 7 and 9 leave a remainder of 3 in each case are 158 - 15 = 143 Hence, option 3.

Volume of the cube = 125 m³
 Length of each side of the cube = 5 m.  Area of each face of the cube = 5 x 5 = 25 m²

3 faces are covered fully by handmade paper. 2 faces are covered with 4 strips of each paper, each strip being of (1/8)th of the area of the cube.
Therefore, each type of paper covers half of each of the 2 faces. Another face is wrapped half by each paper.

Faces covered with handmade paper = 3 + 0.5 + 0.5 + 0.5 = 4.5. Area covered with handmade paper = 4.5 x 25 - Faces covered with wrapping paper = 6 - 4.5 = 1.5.  Area covered with wrapping paper = 1.5 x 25. Required product of the areas = 4.5 x 25 x 1.5 x 25 = 4218.75 m2 
Hence, option 3.

Let the CP of the TV be Rs. x; SP of TV = Rs. 1.2x.

Since the cost price of the TV exceeds the selling price of the radio by Rs. 100, SP of radio = Rs. (x - 100). Since the shopkeeper incurs a loss of 25% on the radio transaction, CP of the radio = (x - 100)/0.75 = Rs. 1.33(x - 100).

Total CP = x + 1.33(x - 100) = Rs. (2.33x - 133) and total SP = 1.2x + (x - 100) = Rs. (2.2x - 100). Since the value of x is not known, the net profit/loss cannot be found.
Hence, option 5.  

Let the centre O of the circle meet the chord AD at point P as shown below. 

Hence, OP = 6 cm and OA = radius of circle = 10 cm

Since ABCDEF is a regular hexagon, AP = PD = BC = BP = 8 cm Let O meet BC at Q such that OQ = x cm.
Hence, in ΔPBQ, BP² = BQ² + PQ²

Now, BQ = BC/2 = 4 cm and PQ - PO + OQ = (x + 6) cm

Let the average weight of the students in M1, M2 and M3 be a, b and c kg respectively.
Since the student leaving M1 has decreased the average weight of the group, this student is heavier than the average weight.
Now, note that if there are x objects and the average value decreases by p, the total value should have decreased by xp.
Since there are 19 students now in M1, and the average weight has reduced by 2 kg, the total weight of the group should have reduced by 2 x 19 = 38 kg.
So, the weight of the student who left M1 would be (a + 38) kg.
Similarly, the student who left M3 would be heavier than the average weight of M3.

Since there are 39 students left in M3 and the average weight of this group reduces by 2 kg, the weight of the student who left M3 = (c + 78) kg.
Both these students joined M2, thereby increasing its average weight by 3 kg. 

Since the dealer has sold the mixture at cost, C.R of mixture = S.R of mixture = Rs. 46 per kg Since the cost of cheese B (Rs. 50 per kg) is more than the cost of the mixture, the other cheese should cost less than the mixture.
So, cheese A should cost less than Rs. 46 per kg.
Hence, options 4 and 5 can be eliminated.
Now, the cost of cheese A can be found, as shown below:

Thus, cheese A costs Rs. 44 per kg.
Hence, option 2

6 = 1 x 2 x 3 = 1 x 1 x 6 So, n can take 9 possible values: 123, 132, 231, 213, 312, 321, 116, 161 and 611.
Now, if the digits are 1, 2 and 3, the sum is 6 while if the digits are 1, 1 and 6, the sum is 8.
Hence, options 1 and 2 can be directly eliminated.
Now, factorise each number.
If a number n is factorised as x its number of factors = (1 + 1)(1 + 1) = (2)(2) = 4.
Hence, if any of the given numbers can be written as a product of exactly two prime factors, it cannot be the value of n (as it will have only 4 factors). 123 = 3 x 41 i.e. rejected.

132 = 2² x 3 x 11 i.e. no. of factors = (2 + 1)(1 + 1)(1 + 1) = 12.

231 = 3 x 7 x 11 i.e. no. of factors = (1 + 1)(1 + 1)(1 + 1) = 8.

213 = 3 x 41 i.e. rejected. 312 = 2³ x 3 x13 i.e. no. of factors = (3 + 1)(1 + 1)(1 + 1) = 16. Hence, 312 is a possible value of n.

321 = 3 x 107 i.e. rejected. 116 = 2² x 29 i.e. no. of factors = (2 + 1)(1 + 1) = 6.

161 = 7 x 23 i.e. rejected 611 = 13 x 47 i.e. rejected. Hence, the only possible value of n is 312 and sum of digits of 312 = 3 + 1 + 2 = 6 Hence, option 4.

Let us divide 14 people into group of 6 and 8 people respectively. 

Within the two groups, number of ways to arrange 6 and 8 around the table 5! and 7! respectively.

There are 486 cyclists at Churchgate. So 486 cyclists left Dadar. Let there be x cyclists before Dadar.
Since half of the cyclists left, there were x/2 cyclists remaining. These were joined by 1/5 of remaining cyclists.

This has to be equal to 486. x = 810
Using the same logic, we have the number of cyclists before Bandra to be 1350, before Santacruz to be 2250, before Andheri to be 3750 and before Malad to be 6250.
Thus total cyclists leaving Borivali is 6250.

Hence, option 1.

Case 1, if AB = BC 

Let Shyam and Ram originally have s and r chocolates respectively.
When Shyam gives 5 chocolates to Ram; s - 5 = r + 5.

s - r = 10 ... (i) When Ram gives 5 chocolates to Shyam; s + 5 = 2(r - 5). So 2r - s = 15 ... (ii) Adding (i) and (ii); r = 25 Hence, option 4.

Let common difference of the AP be d.
We have a₁ = a₃ - 2d = 4 - 2d.

We also have a₁ = a_{N - 2} - ( N - 3 )d = 9 - d( N - 3) = 9 - N d + 3d Equating these two values of a₁, we get,

4 - 2d = 9 -Nd + 3d, or (N - 5)d = 5.

Since both N and d are positive, we must have N - 5 = 1 and d = 5 or N - 5 = 5 and d = 1.
Thus, either N = 6 and d = 5 or N = 10 and d = 1.
In the first case, since d = 5, a₁ = 4 - 2d = 4-2*5 = -6.
But it is given in the question that the terms are all positive integers. So this case is not possible.
In the second case, a_N + 2N + 2 = a₁+ (N - 1)d + 2N + 2 = 4 - 2 x 1 + 9 x 1 + 2 x 10 + 2 = 33.

So value of the given expression = 33 Hence, option 3.

Since the two lines have the same slope, they are parallel to each other. Hence, the largest circle drawn between these lines touches both these lines. Hence, distance between these lines = diameter of the circle For two parallel lines with equations ax + by + c₁ = 0 and ax + by + C₂ = 0,

Hence, option 2.

When a coin is tossed, there are two equally likely outcomes possible, head and tail.
Hence, number o f possible outcomes when five coins are tossed = 2 x 2 x 2 x 2 x 2 = 2⁵ = 32. Even number of heads and odd number of tails gives three cases: 0 heads and 5 tails. This is like arranging a five-letter word where all the letters are same. This can be done in only 1 way. 2 heads and 3 tails. This is like arranging a five-letter word where one letter is present twice and the other is present thrice. 

4 heads and 1 tail.

This is again like arranging a five-letter word where one letter is present four times and the other is present just once.

Hence, total ways of getting even number of heads and odd number of tails = 10 + 5 + 1 = 16

Hence, option 1.

m( 0) = 4(0)³ - 5(0)² + 2(0) +13 = 13.

m(0) < k + 5. 13 < k + 5. k > 8

Hence, options 1 and 4 can be eliminated. m(1) = 4(1)³ - 5(1)² + 2(1) + 13 = 4 - 5 + 2 + 13 = 14.

Since m(1) > k - 15

14 > k - 15
Therefore k < 29 i.e. 8 < k< 29. Among the options, only k = 10 satisfies this condition 
Hence, option 3.

The total amount with the students in the class is Rs. 35x. If 4 students with an average amount of Rs.15 leave the group, the average amount now left would be the same as the amount if 2 students with an average amount of Rs. 90 join the group.

Thus, the class has 20 students.
Hence, option 3.

The given data can be represented as shown below: 

The given equation can be written as: 

Since x is real, this case is not possible.
Thus, only one real value of x satisfies the given equation.
Hence, option 1.

When the hour hand and minute hand are together, the angle between the two hands is 0°.

The hands meet again between 7 a.m. and 8 a.m.
Here, h = 7 and θ = 0.

Without losing generality we can assume that shopkeeper has 100 articles with marked price of 100.
Now, for first 70 articles he offered discount of Rs. 40 on each of the articles.
Hence selling price of these 70 articles will be Rs. 60 each. Total selling price of these 70 articles will be, 70 x 60 = 4200 And he offers additional 30% discount on remaining articles.
Hence selling price of these 30 articles will be 70% of 60 i.e. Rs. 42 each. Total selling price of these 30 articles will be, 30 x 42 = 1260. Total selling price of 100 articles is, 1260 + 4200 = 5460.
This is 20% more than cost price. Hence,

Note that ( x - 2), ( x - 1), x and (x + 1) are four consecutive integers.
The magnitude of no set of four consecutive integers adds up to 7.
Hence, no integral values of x satisfy the given equation. Hence, option 1.

Only students who passed could apply for jobs and all such students applied for jobs.
Number of students applying for a job in ‘Others’ = (0.33 x 800) + (0.12 x 650) + (0.13 x 1100) + (0.28 x 1200) + (0.16 x 1000) = 264 + 78 + 143 + 336+ 160
= 981

Since the number of students applying for Systems jobs has to be mimimized, the number of students applying for HR should be the highest possible.
The number of students applying for HR jobs was the same each year.
Hence, this number is maximized when there are 78 applicants for HR each year (as number of students each year is equal and there cannot be more than 78 applicants in 2009).
Maximum number of students applying for an HR job = 78 x 5 = 390 Minimum number of students applying for a Systems job = 981 - 390 = 591 
Hence, option 2.

In 2009, students applying to Nefertiti = 25% of 48% of 650 Also, students selected in Nefertiti = 50% of 25% of 48% of 650 = 0.5 x 0.25 x 0.48 x 650
= 39
Each got a salary that was 10% above the starting salary paid in Marketing.
Total monthly salary outflow for Nefertiti = 39 x 1.1 x 6390 = Rs. 2,74131. 
Hence, option 3.