In a certain shipment of 160 new buses,²⁄₅ of buses are equipped with emergency exits and ³⁄₈ are equipped with collapsible doors. 59 buses are equipped neither with an emergency exit nor with a collapsible door. How many buses are equipped with both?

A. 69
B. 51
C. 42
D. 23
E. 27

Solution

160 - 59 = 101 busses are equipped either with an emergency exit or a collapsible door or both.

Let n(E) denote buses with emergency exists and n(C) buses denote with collapsible doors.

∴ n(E) = ²⁄₅ x 160 = 64 and n(C) = ³⁄₈ × 160 = 60

Also n(E - C) = n(E) + n(C)- n(E - C)

∴ 101 = 64 + 60 - n(E - C)

∴ n(E-C) = 124-101 = 23

The average (arithmetic mean) of a set of 17 numbers which includes 67 is A. If 67 is removed from the set and three numbers -51, 168 and 121 are added to the set, find the average of the new set of numbers in terms of A?

A. 17A+ 1
B. $\frac{17A+ 1}{18}$
C. $\frac{17}{19}$A+ 1
D. $\frac{17}{19}$A+ 9
E. $\frac{17A + 53}{19}$

Solution

Total of 17 numbers is 17A.

∴ 17 A - 67 - 51 + 168 + 121 is the total of 19 numbers.

∴ 17A + 171 is the total of 19 numbers.

∴ Average =$\frac{17A+171}{9}=\frac{17}{19}A+9$

X is the sum of 5 consecutive integers, the greatest of which is S. In terms of X, which of the following is the sum of five consecutive integers least of which is S?

A. x + 10
B. x - 10
C. x - 20
D. x + 30
E. x + 20

Solution

The greatest integer is S

∴ Five numbers are S-4, S-3, 3-2, S-1, S respectively.

∴ X = (S - 4) + (S - 3) + (S - 2) + (S - 1) + S

∴ 5S = X + 10

Now, in the second case, least integer is S

∴ Five numbers are S, S + 1, S + 2, S + 3, S + 4

Total will be 5S +10 But 5S = X + 10

∴ Required sum is X + 10 + 10 = X + 20.

At the beginning of a week, Erika had $x. During the first 3 days of the week she spent a total of ‘p’ dollars. For the 2 days following that, she spent ‘q’ dollars on each day and for the last 2 days, she spent t dollars per day. Finally she was left with $156. How many dollars did she spend on the last day of week?

A. $\frac{x - 2p - 2q -156}{7}$
B. $x-\frac{2p}{3}-q-156$
C. $\frac{7x - P - 2q -156}{2}$
D. $\frac{x - P - 2q -156}{2}$
E. $\frac{x - P - 2q -156}{7}$

Solution

On Monday, Tuesday and Wednesday, Erika spends 'p' dollars in total. On Thursday and Friday, she spends 'q' dollars on each day. On Saturday and Sunday she spent 'r' dollars each day.

∴ Total money spent in the week = p + 2q + 2r

∴ x - p - 2q - 156 = 2r ∴$\frac{x- p-2q -156}{2}=r$

M = (2²×4²) + (3² × 25¹ × 34^²⁄₃)^¹⁄₂ + 2

What is the average of the distinct prime factors of M?

A. 12.5
B. 11
C. 15
D. 8
E. 14

Solution

M = (2²×4²) + (3² × 25¹ × 34^²⁄₃)^¹⁄₂ + 2

∴ M = 64 + 105 + 2 = 171

∴ M = 3 × 57 = 3 × 3 × 19

∴ The average (arithmetic mean) = $\frac{22}{2}$ = 11

Total sales for business in a certain year = $2,40,000. The sales in June and July were 1.5 times and 2 times the monthly average. Find the average sales for these 2 months in $.

A. 30,000
B. 35,000
C. 25,000
D. 45,000
E. 37,500

Solution

The monthly average =$\frac{2,40,000}{12}$= $20,000

Sales in June and July will be $30,000 and $40,000 respectively.

∴ Average of June and July will be $35,000.

Mark and Maria stay next to each other and study in the same school. Mark and Maria walk towards their school at speeds of 7 kmph and 10.5 kmph respectively. The faster of the two reaches the school first, turns around and starts walking back. They meet 400 m away from home. What is the distance between their residences and school?

A. 250 m
B. 300 m
C. 450 m
D. 500 m
E. 600 m

Solution

Ratio of speed A Marks and Maria is 7: 10.5 i e. 2 : 3

Maria reaches school, returns and meets Mark. 400 meters away from home. It means that Mark has walked only 400 meters. And applying the ratio of speed Maria must have walked ³⁄₂ × 400 = 600 meters.

∴ Total distance crossed by Mark and Maria is 400 + 600 = 1000 meters. Which is twice the distance of school.

∴ Distance between house and school is 500 meters.

Sam D’Souza died leaving behind him an estate of $37,000. According to his will, his three sons Tom, Dick and Harry must get the same amount from his estate and his previous gifts. If he had previously given Tom, Dick and Harry $17,000, $11,000 and $16,000 respectively, how much must Dick get from the estate?

A. $18,000
B. $16,000
C. $15,000
D. $12,000
E. $12,333

Solution

Adding the individual amounts of the gifts and that of the estate, we have

$37000 + $17000 + $11000 + $16000 = $81000

∴ Each son must get$\frac{\$ 81000}{3}=\$ 27000$

∴ Dick must get $27000 - $11000 = $16000 from the estate.

Solution A is 40% of liquid x and 60% of liquid y. 2 kg of liquid y evaporates from 6 kg of solution A and 4 kg of solution A is added to remaining 4 kg of liquid, what % of this new solution is liquid X?

A. 40%
B. 25%
C. 60%
D. 50%
E. 75%

Solution

∴ Finally the 8 kg. solution A 4 kg. liquid X and 4 kg. Liquid Y

Find the square root of $\frac{\left ( 3\frac{1}{2} \right )^{4}-\left ( 2\frac{1}{3} \right )^{4}}{\left ( 3\frac{1}{2} \right )^{2}-\left ( 2\frac{1}{3} \right )^{2}}$

A. 7
B. $\frac{7}{6}\sqrt{11}$
C. $\frac{7}{6}\sqrt{13}$
D. 6
E. $\frac{7}{6}$

Solution

Let $\left ( 3\frac{1}{2} \right )^{2}$ = x and $\left ( 2\frac{1}{3} \right )^{2}$ = y

∴ Given expression =$\frac{x^{2}-y^{2}}{x-y}= x + y$

=$\left ( 3\frac{1}{2} \right )^{^{2}}+\left ( 2\frac{1}{3} \right )^{2}=\left ( \frac{7}{2} \right )^{2}+\left ( \frac{7}{3} \right )^{7}$

=$\frac{49}{4}+\frac{49}{9}=49\left [ \frac{1}{4}+\frac{1}{9} \right ]$

= 49$\left [ \frac{13}{36} \right ]$

∴ Square root of 49$\left [ \frac{13}{36} \right ]=\frac{7}{6}\sqrt{13}$

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