Top 100 questions on QA & DI

Let the speed of A, B and C be 5v, v and v respectively. 

Let the circumference of the track be 12vt. 

Instance when A meets: 

B – 3t, 6t, 9t, 12t, 15t, 18t .... 

C – 2t, 4t, 6t, 8t, 10t, 12t .... 

In every 6t units of time A meets B twice while A meets C thrice. 

So in 36t A would give: B – 2 × 6 = 12 cards C – 3 × 6 = 18 cards. (total 30) Now, at 38t A gives C a card (his 31st) at 39t A gives B a card (his 32nd) and at 40t A gives C a card (his last) 

So the required difference = (18 + 2) – (12 + 1) = 7

S = 1/100 + 1/101 + 1/102  .... 1/998 + 1/999 + 1/1000 ,  then 

In the first 450 terms, the sum of the two numbers in the denominator is constant. Thus, their product will be maximum when they are equal

Let the co-ordinates of A and B be (0,0) and (a,0) respectively.

Let the co-ordinates of point C be (h, k).

Given:

2AC = BC

The equation written above is that of a circle with center (say O) at  (-a/3 – 0)  and radius  2a/3, The center O  (-a/3 – 0)  divides AB externally such that

4AO = BO

a₁ = 1 

a₂ = 2 

a₃ = 1 (1) = 1 

a₄ = 2 (0) = 0 

a₅ = 1 (– 1) = – 1 

a₆ = 0 (– 2) = 0 

a₇ = – 1 (– 1) = 1 

a₈ = 0 (0) = 0 

a₉ = 1 (– 1) = – 1 

After a₁ and a₂, there is a cyclicity of ‘4’ in the remaining terms a₁₀₀₀ = a_{2 + 4 × 249 + 2} = 0.​

The minute-hand of a normal clock covers 360/ 60 =6° per minute. The hour-hand of a normal clock covers 30/60 =1°/ 2 per minute. So once they are together, in every minute the minute hand gains 6 − 1/2 =11°/2  over the hour hand. 

So, time between two meetings = 360/11/2 = 720/11 minutes. 

So, in any clock the hour-hand and the minute-hand meet after every 720/11 minutes. 

If 60 minutes have passed in a normal clock then time passed in the faulty clock is 70 minutes. 

If 24 hrs (or 24 × 60 minutes) have passed in a normal clock then time passed in the faulty clock must be 24 × 70 = 1680 minutes. 

Number of times the hands meet – 

If Brazil never gets a lead over India, the first goal of the match should be scored by India. At this stage India leads by 1 – 0, and the only way in which Brazil can take a lead is by scoring the next two goals i.e. a scoring pattern like IBBIII. In all other cases Brazil would never be able to lead India. 

Total cases where India scores the first goal 5!/3!2! =10 

Different scoring patterns possible = 10 – 1 = 9

Now, three cases arise: 

Case–1 

P(x) = 1 and Q(x) may be anything. 

∴  (x−7) = 1 or x = 8 But, for x = 8, Q(x) is not defined. 

Case–2 

P(x) = – 1 and Q(x) is an even exponent. (x – 7) = – 1 or x = 6 For x = 6, Q(x) = – 4, which satisfies the given equation.

Case–3 

Q(x) = 0 and P(x) 0

⇒ x = 7 or x = 22 But, for x = 7, P(x) = 0, for which the given equation is not defined.

When N = 1 M = 1 × 6 N = 2 M = 6 × 11 N = 200 M = 996 × 1001

Construction: Extend BC to cut the circle at E and join AE.

∆AEB is right angled at E. (Angle in a semicircle is a right angle).

Hence, both AE and OD are perpendicular to EB. By AA property:

∆ AEB ~ âˆ†ODB 

Hence, OB/AB = OD/AE or OD/AE = 1/2 

⇒ AE = 2OD =  3cm.

In right ∆AEC :

 ∠ACE = 180^o −∠ACB = 180^o − 120^o = 60​^o

As per the information given in the question, we can conclude that

⇒ 0.05c + 0.15d = 0.05b + 0.15a

⇒ c + 3d= b + 3a

The ratio 5 : 2 : 7 : 3 does not satisfy the given relation.

3y² = x² – 1376

As we can see L.H.S. is definitely a multiple of 3 and in R.H.S. 1376 leaves a remainder of 2 when divided by 3.

There are three possibilities for x in R.H.S:

(i) If x is multiple of 3, so is x², and R.H.S. will leave a remainder of 1 when divided by 3.

(ii) If x is of the form 3m + 1, x² will be of the form 3n + 1 and R.H.S will leave a remainder of 2. m, n ∈ N 

(iii) If x is of the form 3m + 2, x² will be of the form 3n + 1 and R.H.S. will leave a remainder of 2. m, n ∈ N 

So R.H.S. can never be a multiple of 3, while L.H.S. is always a multiple of 3. Hence no real solution exists.

f(0) > 0 and f(1) < 0 implies that one root for f(x) = 0 lies between x = 0 and x = 1.

f(6) + f(8) = 0 implies that f(6) and f(8) are of opposite sign but same absolute value. Hence another root for f(x) = 0 must lie between x = 6 and x = 8. As f(1) < 0, f(6) must also be less than zero, otherwise we’ll have more than 2 roots for f(x) = 0.

Hence f(8) > 0 and f(6) < 0.

Further f(7).f(9) > 0 implies that both f(7) and f(9) are greater than zero.

So the second root for f(x) = 0 must lie between x = 6 and x = 7.

So f(x) would look like :​

As f(1), f(2) and f(3) are less than zero, f(1).f(2).f(3) < 0 is true.

As f(3), f(5) < 0 and f(7), f(9) > 0, f(3).f(5).f(7).f(9) > 0 is true.

As f(7), f(8) > 0, f(7).f(8) < 0 is false.

f(0), f(9), f(10) > 0 and f(1) < 0, but since we don’t know the magnitude of any of these four we cannot judge if f(0) + f(1) + f(9) + f(10) is greater than zero or not.​

The maximum total work done for three consecutive days will be when A, B and C work in distinct pairs (for example ‘A and B’, ‘B and C’ and ‘C and A’). Hence on these three days fraction of total work done will be

The remaining 4/120 fraction of work can be done in least amount of time if A and B work together on the 7^th day.

A and B together can do 1/8 +1/12 = 5/24 + = fraction of work in a day.

So, they will complete the 4/120 fraction of work in

4/120 ÷ 5/24  = 4/25 day

By Pythagoras theorem:

Since diagonals of a rectangle are equal and bisect each other:

AG = DG = AC/2  = a

Hence, ∆DGA is equilateral and since AF is perpendicular to the base DG we can say that AF must divide ∆DGA into two equal halves.

Area of equilateral  âˆ†DGA = √3/4 a² square units

From Statement A:

This statement is clearly insufficient. The answer can given only when exact values of a, b and K are known.

For example let (a, b) = (4, 2) or a/b = 2 

Case 1: K = 0

(a + K)/(b + K) = a/b

Case 2: K = 2

(a + K)/(b + K) = 6/4 = 3/2 > a/b

Case 3: K = –1

(a + K)/(b + K) = 3/1 = 3 > a/b

From Statement B:

As explained above for Statement A, this statement is also insufficient to answer without knowing the exact values of a, b and K.

Combining Statement A and Statement B:

Even the two statments combined cannot give a unique answer.

If f(0) is non-negative, roots of f(x) = 0 cannot be of opposite signs.

So the possibilities for the two roots are (0, 3), (0, 2), (0, 1), (1, 2), (1, 3), (2, 3), (0, –3), (0, –2), (0, –1), (–1, –2), (–1, –3) and (–2, –3).

Hence the required answer is 12.

Let the quantity of total solution be 800 units. So, quantity of milk is 500 units and water is 300 units. Now x percent of 800 = 8x units mixture is removed. The quantity (in units) of milk and water removed is 5x and 3x respectively.

Final concentration of milk (in percentage) is:

(500 − 5x)/800 × 100  

So, 30 < (500 − 5x)/8 < 50

⇒  240 < 500 − 5x < 400

⇒ − 260 < −5x < −100

⇒ 20 < x < 52.

Distance of origin (0, 0) from the line 3y – 4x – 15 = 0:

Let the new lines drawn parallel to 3y – 4x – 15 = 0
be L₁ and L₂.
Distance of L₁ from origin = 3 + 3 = 6 units
Distance of L₂ from origin = 3 – 3 = 0 units
The circle x² + y² = 25 has a radius of 5 units.
Hence line segment of L₁ lying inside the circle will be
of zero length (L₁ does not cut the circle).
Chord cut by L₂ will be diameter = 10 units

We can see that the difference between the divisor and the respective remainder is the same in each division i.e.

2 – 1 = 4 – 3 = 6 – 5 = 8 – 7 = 1

Hence the general form of such numbers will be LCM(2, 4, 6 and 8).K – 1 = 24K – 1, where ‘K’ is any natural number.

Hence the numbers are 23, 23 + 24, 23 + 2 × 24, ......, 23 + 40 × 24 A total of 41 such numbers are there between 0 and 1000.

a¹⁵ + a¹⁶ + a¹⁷ + ....... +a⁵⁰

Sum = a¹⁵ {1 + a + a2 + ------- + a³⁵}

Since a is the root of equation x⁵ – 1 = 0,​

a⁵ – 1 = 0 ⇒ a⁵ = 1

Let’s assume that ‘I’ denotes the integral part and ‘F’ denotes the fraction part of   (5 + √19)ⁿ.

⇒ I + F = 5ⁿ = ⁿC₁.5^n-1.√19 +ⁿC₂.5^n-2.19 + ⁿC₃.5^n-3.19√19 + ........​ 

Now​ ( 5 − √19 )ⁿ  is a proper fraction as ( 5 − √19 ) < 1

Let’s assume that F′ = ( 5 − √19 )ⁿ

⇒ F′ = 5n − ⁿC₁.5^n-1.√19 + ⁿC₂.5^n-2.19 + ⁿC₃.5^n-3.19√19 + ........​ 

Hence,​

I + F + F′ = 2 [ 5ⁿ + ⁿC₂.5^n-2.19 + ⁿC₄.5^n-4.19²+ ........​ ]

Evidently, F + F = 1 ′ as the sum of two proper fractions is always less than 2 and greater than 0. Also the right hand side is an integer and hence F + F′ should also
be an integer.

Therefore I + 1 is an even number and I is an odd number.

Alternative method:

Putting n = 1, we get 5+ √19 whose integral part is 9.
Putting n = 2, we get 25 + √19 + 10 19 whose integral
part is 25 + 19 + 43 which is again an odd number.
Now, through the options it can be judged that the
greatest integer must always be an odd number.

Let A involves 30 units, B involves 15 units and C involves 20 units of work.
One welder does 3 units of A and 1 unit of C in a day.
One blacksmith does 1 unit of B and 2 units of C in a day.

Two welders would take 5 days to complete A. They finish A at the end of day 5.
On day 6 they finish 2 units of C.
Three blacksmiths finish B at the end of day 6.
After days 6, two welders and three blacksmiths together would take 18/8 days to finish the rest of C.
So in all it takes 33/4 days to complete all the jobs.

∠BAD = 90°

Extend BA and CD to meet at E.

∆EBC becomes an equilateral triangle.

So BE = 4 cm

In ∆ADE, ∠ADE  = 30°

tan30° = AE/2 = 1/√3

⇒ AE = 2/√3 cm

∴​ AB = BE − AE = 4 − 2/√3 = (4√3 − 2)/√3  cm

Discount to Saral 20 + 25 –(20 × 25)/100 = 40%

Discount to Himanshu = 40 + 40 – (40 × 40)/100 = 64% 

Let a, b, c be the number of units sold at 20%, 40% and 64% respectively

⇒ (a × 20)/100 + (b × 40)/100 + (c × 64)/100 = {(a+b+c)50}/100

⇒ 7c/50 – b/10 – 3a/10 = 0

⇒ 7c – 5b – 15a = 0

So ‘c’ has to be a multiple of 5. Values of a, b, c satisfying the equation are (2, 1, 5); (1, 4, 5); (4, 2, 10); (3, 5, 10); (2, 8, 10).

Hence the possible values for x are 8, 10, 16, 18 and 20.​

Area of ∆DEC = Area of ∆DBC ……..( ∆'s between same parallel line and same base)

= 1/2 × Area of parallelogram ABCD​

Area of ∆EFG = 1/2 × Area of parallelogram EBCF​

= 1/2 × 3/5 × Area of parallelogram ABCD​

Case 1:
All four balls in the same box: 1 way
Since all the boxes are identical.

Case 2:
Two balls in 1 box and the remaining two in another
box:

⁴C₂/2! = 3 ways

It is divided by 2! because the boxes are identical.

Case 3:
Two balls in a box and one ball in each of the remaining two boxes:

⁴C₂ = 6 ways

Note: We need not select a box for the remaining two
balls as they will go one each in the remaining two
boxes in one way only.

Case 4:
Three balls in one box and one remaining ball in another box:​

⁴C₃ = 4 ways
Since there is only one way of selecting the other box as the boxes are identical.
Total ways = 1 + 3 + 6 + 4 = 14

Let the number of shares sold by Sajid and Hasan on Day 1 be 36x each.

Investment made by Sajid
= 375 × 18x + 250 × 18x = Rs. 11,250x.
Profit made by Sajid
= 750 × 18x + 625 × 18x – 375 × 18x – 250 × 18x = Rs. 13,500x.
Sajid’s Margin on Day 1 ≈ 54.5%.

Investment made by Hasan
= 225 × 36x = Rs. 8,100x.
Profit made by Hasan
= 375 × 16x + 300 × 20x – 225 × 36x = Rs. 3,900x. Hasan’s Margin on Day 1 = 32.5%.

For Hasan:
As the share price at 11:00 a.m. and 12:00 noon was Rs. 500 and Rs. 400 respectively, the number of shares sold by Hasan at 11:00 a.m. and 12:00 noon must be in the ratio 4 : 5 respectively. Let the number of shares sold by Hasan at 11:00 a.m. and 12:00 noon be 4x and 5x respectively.
Total sales amount = 500 × 4x + 400 × 5x = Rs. 4,000x Total investment in purchase = 750 × 9x = Rs. 6,750x

Margin (loss) = (6750 − 4000)/4000 = 68.75%

For Sajid:
Let the total number of shares sold by Sajid at 9 a.m. and 10 a.m. be 2y. Total sales amount = 200 × y + 300 × y = Rs. 500y
Total investment in purchase = 500 × y + 625y = Rs. 1,125y

Margin (loss) = (1125 − 500)500 = 125% Required ratio = 68.75 : 125 = 11 : 20

Player 1 and Player 2 of Team 4 have better averages than Player 2 of Team 3.

Runs scored by Player 1 and Player 2 of Team 4 = 13600 + 10800 = 24400

Matches played by them = 340 + 240 = 580

Average score = 24400/580 = 42.07

FOR DIRECTIONS FOR QUESTIONS 75 to 77 :

Aamir is either A or B. Accordingly, Abhishek would be either B or A. Therefore, Amitabh and Akshay are disguised as C and D. As the maximum possible marks in each Mock Cat were 75, the total marks obtained by C are definitely more than the total marks obtained by D. Therefore, C is Amitabh and D is Akshay.

Let us assume that Abhishek is disguised as B and the marks scored by Abhishek in Mock 1 are x.

∴  x + 87 > 138 or x > 51 which would mean that Abhishek could not score the lowest in Mock 1. Hence, the assumption is incorrect and so Abhishek must be disguised as A and Akshay as B​

Let’s assume that Akshay, who is disguised as D, score p marks in Mock 3.

So his total score d = 63 + p

As we know that d < c

63 + p < 138

or p < 75

All the scores from –25 to +75 are possible in any of the tests, except 70, 73 and 74.

Therefore, different scores possible for Akshay = 101 – 3 – 1 = 97

Abhishek is disguised as A. Let’s assume that Abhishek scores q marks in Mock 2. As Aamir scored the highest in Mock 2, q < 62 and 47 + 72 + q > 51 + 59 + 28.

∴ 19 < q < 62

Abhishek’s score is a multiple of 6.

∴ 119 + q needs to be a multiple of 6.

∴ q = 6K + 1, where K is a whole number.

19 < 6K + 1 < 62

∴ 4 ≤ K ≤ 10 Therefore, 7 different values are possible for a.

FOR DIRECTIONS FOR QUESTIONS 78 to 80 :

Let the fee (in Rs. lakhs) demanded by Careena, Carishma, Cashmira and Catrina be a, b, c and d (not necessarily in that order).
Let the fee (in Rs. lakhs) demanded by Coena be e.
Also, either Z or Y must have demanded Rs. 1123 lakhs as the prices demanded by X, U and V are already given.

Let e = 1123.
a + b = c + d.
It can be seen that 2397 – 2379 = 18
and 18 + 1213 = 1231.
So, 1231 + 2379 = 2397 + 1213.
Also, 1201 < 1231 < 2288.
So, 1231 is one such acceptable value.
Coena must have demanded the fee of Rs. 1123 lakhs and so she must be either Z or Y (as concluded above).​

The following cases are possible:​

Note:
e cannot take any value other than 1123.
E.g. let e = 2397.
In such a case no equation of the type ‘a + b = c + d’ is possible, which satisfies the given conditions. A possible equation is ‘1123 + 2289 = 1213 + 2379’ but since 2289 > 2288, it cannot be taken.​

FOR DIRECTIONS FOR QUESTIONS 81 to 83 :

Let the number of fit sportsmen who play Hockey be x. From statement (ii), the number of unfit sportsmen who play Hockey is also x. From statements (v) and (vii), the number of fit sportsmen who play Cricket is 1250 and the number of unfit sportsmen who play Cricket is 0. From statement (i), we can deduce that x is equal to 2500. From statement (iv), the number of sportsmen who play Hockey for Under-19 team is 4000 and the number of sportsmen who play Hockey for Above-19 team is 1000. Further analysis leads to the tables given below.​

The total number of fit sportsmen who play for Above-19 team is 1000.​

DIRECTIONS FOR QUESTIONS 84 to 86 :

The table given below shows the grades received by the students in different subjects.​

DIRECTIONS FOR QUESTIONS 87 to 89 :

Let Interest Payment be x% and Non-Planned Capital Account be y% of Budgeted Expenditure. From pie chart B, x + y = 22% and x – y = 12%. Thus, x = 17 and y = 5.
Now 76% – 19% i.e. 57% of Budgeted Revenue is given as Rs. 285 thousand crores. Thus Budgeted Revenue = Rs. 500 thousand crores.
Budgeted Expenditure = Budgeted Revenue + Rs. 625 thousand crores = Rs. 1125 thousand crores

The difference between Budgeted Expenditure and Budgeted Revenue when expressed as a percentage of Budgeted Revenue

= 625/500 × 100 = 125%

Percentage change for the year 1971 is the highest and the value is