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Amar and Akbar complete the project in minimum time ⇒ Anthony is the slowest person.

Amar and Anthony complete the project in maximum time ⇒ Akbar is the fastest person.

Assume that Amar, Akbar and Anthony complete the project independently in a, b and c months.

∴ in one month, the three completes 1a,1b,1c units of work.

Thus, 1a + 1b = 1/12, 1b + 1c = 1/16, 1a + 1c = 1/24
 
Adding the three equations, we get 2*(1a + 1b + 1c) = 1/12 + 1/16 + 1/24

∴ 1a + 1b + 1c = 3/32
 
∴ 1a + 1b + 1c -1b + 1c = 3/32 - 1/16 = 1/32
 
Amar will take 32 months to complete the project. 

Answer: 32.

OP = OR = 4 cm
 
∴ PB = RB = 4 cm
 
As BC = 10 cm, CR = 10 – 4 = 6 cm
 
∴ CQ = 6 cm
 
Assume that AP = AQ = y cm
 
By Pythagoras theorem,
 
(6 + y)² = 10² + (4 + y)²
 
⇒ 36 + 12y + y² = 100 + 16 + 8y + y²
 
⇒ 4y = 116 – 36 = 80
 
⇒ y = 20
 
A(∆ABC) = ½ × AB × BC = ½ × (20 + 4) × 10 = 120 sq. inches
 
Therefore, the required answer is 120.

Answer: 120.

Let the number of patients admitted in hospital A be ‘x’.
 
∴ the number of patients admitted in hospital B = (x + 21)

By the given condition, 200/x = 152/ (x + 21)  + 3 

∴200/x- 152/(x+21) = 3 

∴ 200x – 152x + 4200 = 3x(x + 21)
 
∴ 48x + 4200 = 3x(x + 21)
 
∴ 16x + 1400 = x² + 21x
 
∴ x² + 5x – 1400 = 0
 
∴ (x – 35)(x + 40) = 0

∴ x = 35

Answer: 35.

Let Neeta, Geeta and Sita earn Rs. n, Rs. g and Rs. s respectively.

By the given conditions,
 
n + g = 6s and s + n = 2g
 
∴ (n + g) – (s + n) = 6s – 2g
 
∴ g – s= 6s –2g
 
∴ 3g = 7s
 
∴ g : s = 7 : 3

If g = 7x, s = 3x. Substituting these values in n + g = 6s, we get n + 7x = 18x or n = 11x
 
Thus Neeta earns the most and Sita earns the least.

Therefore, the required ratio = 11 : 3

Hence, option 1

5 - log₁₀  √1 + x + 4log₁₀ √1 − x = log₁₀ 1/ √1 − x²

5 - log₁₀  √1+ x + 4log₁₀ √1 − x = log₁₀(1/√1 − x * 1/√1 + x)

5 - log₁₀  √1 + x + 4log₁₀ √1 − x  = - log₁₀ √1 − x - log₁₀ √1 + x

4log₁₀ √ 1− x + log₁₀ √1 − x = -5

log₁₀ √1− x = -1

√1− x = 1/10

100x = 99.

Answer: 99

It is known that 3 and 5 are always selected.

i. Either 7 or 8 is definitely selected, but not both.

Consider {1, 2, 4, 6}. Number of possible subsets of {1, 2, 4, 6} are 2⁴ = 16. 

One among 7 and 8 is selected in 2 ways.

Total number of subsets in this case = 2 × 16 = 32

ii. Neither 7 nor 8 is selected.

Consider {1, 2, 4, 6}. Number of possible subsets of {1, 2, 4, 6} are 2⁴ = 16. One of the subsets is an empty set. We will not count this as there would be only 2 selected numbers in this case. 

Total number of subsets in this case = 16 – 1 = 15

Total number of groups = 32 + 15 = 47

Answer: 47.

x₀ = 1, x₁ = 2

x₂ = (1 + x₁)/x₀ = (1 + 2)/1 = 3 

x₃ = (1 + 3)/2 = 2

x₄ = (1 + 2)/3 = 1

x₅ = (1 + 1)/2 = 1

x₆ = (1 + 1)/1 = 2

Thus, (x₀, x₁, x₂, x₃, x₄) = (1, 2, 3, 2, 1). As x₅ = 1 and x₆ = 2, the next three terms must be 3, 2, 1 respectively. 
 
(x₅, x₆, x₇, x₈, x₉) = (1, 2, 3, 2, 1)… and so on 2020 is divisible by 5

Therefore x₂₀₂₀ = 1 and hence, x₂₀₂₁ = 2

Hence, option 2

Given that onion is sold for 5 consecutive months at the rate of Rs 10, 20, 25, 25, and 50 per kg respectively.
 
Let’s consider the amount spend on onions in these five months to be 100, 100, 100, 50, 50.
 
Hence, the number of kgs in these five months should be 10, 5, 4, 2, 1 respectively.
 
The average expense on onions per kg over these five months is given by
 
=  total expense on onions in these five months/ total number of kgs of onions 
 
= 400/22 = 18.181818
 
Therefore, the closest average would be 18. 

Hence, option 1.

The interest accrued during the second year = Rs. 806.25
 
In the third year the interest accrued = 806.25 + interest on 806.25
 
∴ 866.72 = 806.25 + interest on 806.25
 
∴ The interest on 806.25 = 866.72 – 806.25 = 60.47
 
∴ Rate of interest= (60.47×100)/806.25= 7.5% p.a 
 
In the fourth year the interest accrued = 866.72 + interest on 866.72
 
= 866.72+ (7.5×866.72)/100
 
= 931.72 

Hence, option 1.

Let's consider the selling prices of small and large chocolate boxes be ‘x’ and ‘2x’ respectively.
 
Let's also consider the weights of the small and large chocolate boxes to be 100(to make it easy) and y respectively.
 
Given that, the selling price per gram of chocolate in the large box is 12% less than that in the small box.
 
Therefore, x/100 * 0.88 = 2x/y

Hence, option 2
 
y = 200/0.88
 
y ≅ 227

Let the three digit number be (100x + 10y + z), x ≥ 1

The number in reverse order = (100z + 10y + x)

(100x + 10y + z) + 198 = (100z + 10y + x)

∴ 99(z – x) = 198

∴ z – x = 2

So, the difference between the hundreds place digit and the units place digit is 2.

∴ x can take values 1, 2, 3, 4, 5, 6, 7 and accordingly values of z would be 7, 6, 5, 4, 3, 2, 1.

y can take 10 values from 0, 1, 2, …, 9

Therefore, there are 7 × 10 = 70 such numbers.

Answer: 70.

Assume that an apple costs Rs. X, an orange costs Rs. Y and a mango costs Rs. Z.
 
Let there be m number of mangoes in a basket such that the cost of the basket is same as the cost of the other baskets.
 
2X + 4Y + 6Z = X + 4Y + 8Z = 8Y + 7Z = mZ
 
2X + 4Y + 6Z = X + 4Y + 8Z ⇒ X = 2Z
 
X + 4Y + 8Z = 8Y + 7Z ⇒ X = 4Y – Z
 
X = 2Z and X = 4Y – Z ⇒ 3Z = 4Y
 
Now, mZ = 8Y + 7Z = 6Z + 7Z = 13Z
 
Thus, the basket has 13 mangoes.

 Hence, option 1.

Consider |n – 60| < |n – 100| 

For n =(100 + 60)/2 =80,n –60=|n –100|  

For n < 80, |n – 60| < |n – 100| 

|n – 100| < |n – 20|

For n =(100 + 20)/2 =60,n –100=|n –20|  

For n > 60, |n – 100| < |n – 20|

Therefore 60 < n < 80
 
As ‘n’ is an integer, n = 61, 62, 63, ….., 79

There are 19 integers satisfying |n – 60| < |n – 100| < |n – 20|

Hence, option 3.

x² – 4x – 13| = r

x² – 4x – 13 = x² – 4x + 4 – 17 = (x – 2)² – 17

x² – 4x – 13 = r OR x² – 4x – 13 = – r

∴ (x – 2)² – 17 = r OR (x – 2)² – 17 = – r

∴ (x – 2)² = (17 + r) OR (x – 2)² = 17 – r

Consider (x – 2)² = (17 + r)

As r is positive, 17 + r is positive and hence, 

x – 2 = ± 17 + r are the two roots.

Consider (x – 2)² = (17 – r)

As the given equation has three roots, (x – 2)² = (17 – r) must have equal roots. 

∴ 17 – r = 0

∴ r = 17

 Hence, option 1.

The solutions of strength 33% and 17% are mixed and the resultant solution has 21% strength.

Therefore the ratio of the solution from the first bottle to the second bottle = 4 : 12 = 1 : 3

The solution from the second bottle = 3/4 × 800 = 600

Therefore, the solution left in the second bottle = 800 – 600 = 200 cc

Answer: 200.

Let the side of the hexagon be ‘x’ cm.
 
Area of the hexagon = Area of the equilateral triangle of side 12 cm

∴ 6*  √3/4  x² = √3/4*12²
 
= x² = 24

⇒ x = 2 √6

Hence, option 3.

Let there be ‘x’ pens in all. Let the wage of the employee be Rs. W

When he sells 100 pens at Rs. 12 and remaining at Rs. 11, the total selling price 
 
= 1200 + 11(x – 100) = 11x + 100
 
In this case he makes a net profit of Rs. 300.
 
∴ 11x + 100 – (8x + W) = 300
 
∴ 3x – W = 200…(i)
 
When he sells 100 pens at Rs. 12 and remaining at Rs. 9, the total selling price 
 
= 1200 + 9(x – 100) = 9x + 300
 
In this case he makes a net loss of Rs. 300.
 
∴ (8x + W) – (9x + 300) = 300
 
∴ W – x = 600…(ii)

Solving (i) and (ii) simultaneously, we get x = 400

∴ W = 1000

Answer: 1000.

LCM(15, 12, 20) = 60

Assume that the work is of 60 units.

Anu, Vinu and Manu complete 4, 5 and 3 units in one day.

Anu and Vinu work on day 1, day 3, day 5, day 7, …. The two finish 4 + 5 = 9 units per day.

Manu and Vinu work on day 2, day 4, day 6, day 8, …. The two finish 3 + 5 = 8 units per day.

Work done in first two days = 9 + 8 = 17 days.

Hence, in first six days 17 × 3 = 51 units will be completed.

Remaining 9 units will be completed by Anu and Vinu on the 7^th day.

Hence, option 4.

Note that the largest numbers in 1^st, 2^nd and 3^rd groups are 1, 4 and 9 respectively i.e., squares of 1, 2 and 3.

Hence, the largest number in 14^th and 15 groups are 14² = 196 and 15² = 225 respectively.

Thus, 15^th group = {197, 198, 199, …, 225}

There are 225 – 197 + 1 = 29 numbers in this group.

The sum of the numbers in the 15^th group
 
= 29/2* (197 + 225) = 6119

Hence, option 2

The speed of the faster train
 
= 160/12
 
= 40/3 m/s
 
The speed of the other train 
 
= 40/3 m/s  – 6 km/hr

= 40/3- 6× 5 /18
 
= (40-5)/3
 
= 35/3 m/s 

As the trains move in the opposite directions, their relative speed
 
= (40+35)/3
 
= 25 m/s 

∴14 = (160 + length of the other train)/ 25

Solving this, we get the length of the other train = 190 m

Hence, option 4

T is midpoint of CD. Therefore, CT = DT = 1 cm
 
ABCDEF is a regular hexagon. Therefore m∠ABC = 120° 
 
In ∆ABC, AB = BC. 
 
∴m∠BAC = m∠BCA = 30°
 
Consider ∆BJC. As m∠BCJ = 30°, m∠CBJ = 60° and BC = 2 cm
 
JC = Side opposite to 60° = 3 cm 
 
Similarly, AJ = 3 cm
 
∴ AC = 2√3 cm
 
∴m∠ACT = 120 – 30 = 90°
 
In ∆ACT, AC = 2√3 cm and CT = 1 cm
 
AT² = AC² + CT²
 
AT² = (2√3)² + 1²
 
AT = √13 cm

Hence, option 4

x² + 2x – 15 = (x + 5)(x – 3)
 
For x ∈(–5, 3), x² + 2x – 15 < 0

x² – 7x – 18 = (x – 9)(x + 2)
 
For x ∈(–2, 9), x² – 7x – 18 < 0

Note that for x ∈(–2, 3), the numerator as well as the denominator is negative.
 
For (–5 < x < –2) the numerator is negative but the denominator is positive.
 
For (3< x < 9) the numerator is positive but the denominator is negative.

Hence, option 2.