Decision Making (Level Easy)

If a class has x girls and y boys, the total number of students is (x + y). The fraction representing girls is x/(x + y) - the number of girls divided by total students. Option A would give the fraction of boys, and options B and D incorrectly multiply x and y in the denominator.

Using the definition x ^ y = 2x + y, evaluate both sides: 2 ^ a = 2(2) + a = 4 + a, and a ^ 3 = 2(a) + 3 = 2a + 3. Setting them equal: 4 + a = 2a + 3. Solving: 4 - 3 = 2a - a, so 1 = a. Verifying: 2 ^ 1 = 5 and 1 ^ 3 = 5, both equal.

Use the principle of inclusion-exclusion. Total students who studied at least one language = Total - Neither = 15 - 3 = 12. Students who studied both = Latin + Greek - Union = 7 + 8 - 12 = 3. This counts students who studied both subjects exactly once.

Convert gallons to pints step by step. 4 gallons = 4 × 4 = 16 quarts. Since 1 quart = 2 pints, 16 quarts = 16 × 2 = 32 pints. Each bottle holds half a pint, so number of bottles = 32 ÷ 0.5 = 64.

From 9x - 3y = 12, divide by 3: 3x - y = 4. From 3x - 5y = 7. Solving these simultaneously: Subtract second from first: 4y = -3, so y = -3/4. Then 3x = 4 + y = 13/4, x = 13/12. Therefore 6x - 2y = 2(3x - y) = 2(4) = 8.

 Divide given quantity of coffee in 2:1 ratio which is equal to 2/3 and 1/3 1st case---cost for 1 gm coffee=(2/3 gm of 33p per gm)+(1/3 gm of 24p per gm) =(233/3)+(24/3) =22+8=30pcost of 100 gm =3000p=Rs.302nd case---cost for 1 gm coffee=(2/3 gm of 24p per gm)+(1/3 gm of 33p per gm) =(224/3)+(33/3) =16+11=27p cost of 100 gm =2700p=Rs.27so, shop saved 30-27=Rs.3

When 0 < x < 1, squaring makes it smaller: x² < x. Negating makes it negative: -x < 0 < x. Taking reciprocal makes it larger than 1: 1/x > 1 > x. Therefore 1/x is the largest among the given options.

The first sentence asks about combinations, but the actual question and options are about comparing fractions to 1/3. 22/62 ≈ 0.355 > 1/3. 15/46 ≈ 0.326 < 1/3. 2/3 = 0.667 > 1/3. 1 = 1 > 1/3. So only 15/46 is less than 1/3. The dancing groups question appears to be a separate, garbled question.

Let boat speed = b km/hr and stream speed = s km/hr. Downstream: b + s = 40. Upstream: b - s = 14. Adding: 2b = 54, b = 27. Subtracting: 2s = 26, s = 13. Therefore stream speed is 13 km/hr.

Let x = 0.75. Numerator = x³ - 0.001 = 0.75³ - 0.001 = 0.421875 - 0.001 = 0.420875. Denominator = x² - 0.075 + 0.01 = 0.75² - 0.075 + 0.01 = 0.5625 - 0.075 + 0.01 = 0.4975. Result = 0.420875/0.4975 ≈ 0.846. Option A (0.845) is closest.

A takes 8 days, so A's rate = 1/8 work per day. B works 3 times faster, so B's rate = 3/8 work per day. C works 5 times faster, so C's rate = 5/8 work per day. Combined rate = (1+3+5)/8 = 9/8 work per day. Time together = 1 ÷ (9/8) = 8/9 days.

 A company contracts to paint 3 houses.
Mr.Brown paints in 6 days = 1 house
Mr.Brown paints in 1days =1/6 house
Mr.Brown paints in 8days =8/6 =4/3 house
work left = 3-4/3 = 5/3 house
Mr.Black paints in 8 days = 1 house
Mr..Black paints in 1days =1/8 house
Mr..Black paints in 6days =6/8= 3/4 house
work left = 5/3 - 3/4 = 11/12=
.Blue paints in 12 days = 1 house
.Blue paints in 1 day = 1/12 house
Mr. blue will finish 11/12 huse in = (11/12)/12=11 days

Let principal be P and rate be r% per annum. Amount after 4 years = P + (P×r×4)/100 = 1120. Amount after 5 years = P + (P×r×5)/100 = 1200. Subtracting: (P×r)/100 = 80, which is the simple interest for 1 year. Substituting back: P + 320 = 1120, so P = 800. Therefore, Rs. 800 amounts to Rs. 1120 in 4 years and Rs. 1200 in 5 years at simple interest.

Starting with 13 = 13w/(1-w). Divide both sides by 13: 1 = w/(1-w). Multiply both sides by (1-w): 1-w = w. Add w to both sides: 1 = 2w, so w = 1/2. Now calculate (2w)² = (2 × 1/2)² = 1² = 1.

Initially: 60 games played with 30% wins = 18 wins. Let x be consecutive wins needed. Total wins = 18 + x. Total games = 60 + x. We want (18+x)/(60+x) = 50% = 0.5. Solving: 18 + x = 0.5(60 + x) = 30 + 0.5x. So 0.5x = 12, and x = 24. The team needs to win 24 games in a row to reach 50% overall.

Let x = time for each non-math problem, so 2x = time for each math problem. Total: 150×x + 50×2x = 180 minutes. This gives 150x + 100x = 180, so 250x = 180, and x = 0.72 minutes. Time for 50 math problems = 50 × 2x = 100 × 0.72 = 72 minutes. Note: This requires interpreting 'among these questions are 50 mathematical problems' as meaning 150 other + 50 math = 200 total, not 200 + 50.

Original contribution per person = D/M. After 3 men drop out, contribution per person = D/(M-3). Increase = D/(M-3) - D/M = D(M - (M-3))/(M(M-3)) = 3D/(M² - 3M). This gives option D.

Given (a-b)/3.5 = 4/7 and a, b are positive integers. Cross-multiply: 7(a-b) = 4 × 3.5 = 14, so a - b = 2. Since a - b = 2 > 0, we have a > b, which means b < a. Option A is correct.

Let a, b, c be kg of teas A, B, C respectively. We have: a + b + c = 100 (total mixture), b = c (equal quantities), 95a + 100b + 70c = 90 × 100 = 9000 (total value). Substituting c = b: a + 2b = 100, and 95a + 100b + 70b = 9000. So 95a + 170b = 9000. From first equation: a = 100 - 2b. Substituting: 95(100 - 2b) + 170b = 9000. This gives 9500 - 190b + 170b = 9000, so -20b = -500, and b = 25. Then c = 25, and a = 100 - 50 = 50. Answer: 50 kg, 25 kg, 25 kg, which is option B.