RE 2 - Arithmetic Revision Exercise

Let the speed of the escalator be e steps/second.

Since Romesh moves at 4 steps/second, he will take 120/4 = 30 seconds to reach the bottom; and in 30 seconds the escalator will take 30e steps.  
∴ The total number of steps in the escalator = 120 + 30e.

Since Priya moves at 2 steps/second, she will take 100/2 = 50 seconds to reach the bottom; and in 50 seconds the escalator will take 50e steps. 
∴ The total number of steps in the escalator = 100 + 50e.

∴ 120 + 30e = 100 + 50e
⇒ e = 1

∴ The number of steps in the escalator = 120 + 30e = 150.

Hence, option (c).

T₁₁ = a, T₁₀ = a – 2, … , T₆ = a – 10, …, T₁ = a – 20

We know 11 consecutive even natural numbers are in Arithmetic Progression, hence their average will be the middle term i.e., the 6th term here.

∴ x = a – 10

Now, we need to find the average of 15 consecutive natural numbers starting from x. The average will be the 8th number, i.e., x + 7 = a - 10 + 7 = a - 3.

Hence, option (d).

Let the density (kg/litre) of A and B be 3d and 2d kg/litre.

60 kg of A is same as 60/3d = 20/d litres of A
40 kg of B is same as 40/2d = 20/d litres of B

i.e., equal volume of both A and B is mixed.

Initial density of the mixture = $\frac{60+40}{{\displaystyle \frac{20}{d}}+{\displaystyle \frac{20}{d}}}$ = $\frac{5d}{2}$

After t hours,
Weight of A left = 60 – 6t and weight of B left = 40 – 2t
New density of the mixture = $\frac{60-6t+40-2t}{{\displaystyle \frac{60-6t}{3d}}+{\displaystyle \frac{40-2t}{2d}}}$ = $\frac{100-8t}{{\displaystyle \frac{240-18t}{6d}}}$ = $\frac{(100-8t)d}{(40-3t)}$

Now, final density is half of initial density,

⇒ $\frac{(100-8t)d}{(40-3t)}$ = $\frac{4}{5}$ × $\frac{5d}{2}$

⇒ 100 – 8t = 80 – 6t
⇒ 20 = 2t
⇒ t = 10 hours.

Hence, option (d).

Let total cost of Praja’s luggage be Rs. P (without the free allowance)
Let total cost of Raja’s luggage be Rs. R (without the free allowance)
Let free allowance is worth Rs. ‘f’ for a passenger.

Now, 
cost of Praja’s luggage after free allowance = P – f = 2400   …(1)
cost of Raja’s luggage after free allowance = R – f = 1200   …(2)
⇒ P + R – 2f = 3600   …(3)

Had the entire luggage been carried by one person, the cost after free allowance would be
P + R – f = 5400   …(4)

Solving (3) and (4) we get
f = Rs. 1800

⇒ P = 4200 and R = 3000 [from (1) and (2)]
∴ Charge for their entire luggage is in the ratio 7 : 5 which will also be the ratio of the weights of their luggage.

∴ Weight of Praja’s luggage = 7/12 × 60 = 35 kgs.

Hence, option (d).

Let total cost of Praja’s luggage be Rs. P (without the free allowance)
Let total cost of Raja’s luggage be Rs. R (without the free allowance)
Let free allowance is worth Rs. ‘f’ for a passenger.

Now, 
cost of Praja’s luggage after free allowance = P – f = 2400   …(1)
cost of Raja’s luggage after free allowance = R – f = 1200   …(2)
⇒ P + R – 2f = 3600   …(3)

Had the entire luggage been carried by one person, the cost after free allowance would be
P + R – f = 5400   …(4)

Solving (3) and (4) we get
f = Rs. 1800

⇒ P = 4200 and R = 3000 [from (1) and (2)]
∴ Charge for their entire luggage is in the ratio 7 : 5 which will also be the ratio of the weights of their luggage.

Free allowance is worth Rs. 1800

Praja’s total luggage charge without free allowance = Rs. 4200 for 35 kgs.

⇒ Rs. 1800 would be charged for $\frac{1800}{4200}$ × 35 = 15 kgs.

Hence, option (b).

Let the total length of the tunnel be 10x meters.

Let the distance between the train and entry of the tunnel be d meters.
The man is at a distance of 2/5 × 10x = 4x from the entry of the tunnel.

Let the speed of the train be t kmph.

Case 1: Man runs toward the entry of the tunnel
By the time man reaches the entry point, train also reaches there. Hence, the ratio of their speeds should be same as the ratio of their distance travelled.

⇒ $\frac{4x}{d}$ = $\frac{m}{t}$     ...(1)

Case 2: Man runs toward the exit of the tunnel
By the time man reaches the exit point, train also reaches there. Hence, the ratio of their speeds should be same as the ratio of their distance travelled.

⇒ $\frac{4x}{d}$ = $\frac{6x}{d+10x}$

⇒ 2d + 20x = 3d
⇒ d = 20x

From (1) ⇒ $\frac{4x}{20x}$ = $\frac{10}{t}$

⇒ t = 50 kmph.

Hence, option (b).

Amount initially (in paise) = 50x + 100y
Amount finally (in paise) = 100x + 50y

According to the question,
100x + 50y = 1.3 × (50x + 100y)
⇒ 100x + 50y = 65x + 130y
⇒ 35x = 80y

⇒ $\frac{x}{y}$ = $\frac{16}{7}$

⇒ y has to be a multiple of 7.

The only option possible is option (d).

Hence, option (d).

Let the total work to be done = LCM (60, 30, 20, 15, 12) = 60 units.
Efficiency of A = 1 units/day
Efficiency of B = 2 units/day
Efficiency of C = 3 units/day
Efficiency of D = 4 units/day
Efficiency of E = 5 units/day
Efficiency of F = 6 units/day

Since time taken by one group is double the time taken by other group, efficiency of group first should be half the efficiency of second group.

∴ We need to form 2 groups such that the total efficiency of one group is double the total efficiency of another group.

Total efficiency of all 5 = 1 + 2 + 3 + 4 + 5 + 6 = 21
∴ We need to divide people such that one group has combined efficiency of 7 and the other has combined efficiency of 14.

Group 1: A + F
Group 2: B + C + D + E

Group 1: B + E
Group 2: A + C + D + F

Group 1: C + D
Group 2: A + B + C + E + F

∴ 3 such possible combinations of groups are possible.

⇒ 4n + 2 = 14.

Hence, 14.

Jai:
Jai changes speed after every 200 kms.

Time taken to cover first one-third of the distance = 200/10 = 20 hours.
Hence, he changes speed at 10 a.m. + 20 hours = 6 a.m. the following day.

Time taken to cover second one-third of the distance = 200/20 = 10 hours.
Hence, he changes speed at 6 a.m. + 10 hours = 4 p.m. the following day.

Time taken to cover final one-third of the distance = 200/40 = 5 hours.
Hence, he changes speed at 4 p.m. + 5 hours = 9 p.m. the following day.

Vijay:
Let Vijay travels for t hours at each of the three speeds.
∴ 10t + 20t + 30t = 600
⇒ t = 10 hours.
⇒ Vijay takes 30 hours to reach the other end.

Let us consider their positions after 20 hours i.e., at 6 a.m. the following day.

distance travelled by Jai = 10 × 20 = 200 kms.
Distance travelled by Vijay = 10 × 10 + 10 × 20 = 300 kms.
∴ Distance between them at 6 a.m. = 600 – 200 – 300 = 100 kms.

Now, both of will not change their speeds for next 10 hours.

Jai will be travelling at the speed of 20 kmph while Vijay is travelling at speed of 30 kmph.

∴ Time taken for them to meet = $\frac{100}{50}$ = 2 hours.

∴ Time when they meet = 6 a.m + 2 hours = 8 a.m.

Hence, option (c).

Let the number of hours workers work each day be h hours and they complete the work in ‘d’ days. 
Total man hours = h × d
The wage = Rs. x/day
Total payment = xd = Rs. 1347    

New wage = x + x/8 = 9x/8 per day.
Number of hours workers work now = h + h/20 = 21h/20 per day.

∴ Number of days required to complete the work = $\frac{hd}{21h/20}$ = $\frac{20}{21}d$

⇒ Total payment they will now receive = $\frac{9x}{8}$ × $\frac{20d}{21}$ = $\frac{15}{14}$ × xd = $\frac{15}{14}$ × 1347 = Rs. 1443.

Hence, option (c).