28 November 2025 · Shift 2

Number series See how much the series grows at each step: 15-5=10, 35-15=20, 75-35=40. The gaps themselves double every time — $10,20,40$ — so the next gap must be 80. Hence the missing term is 75+80=155. The same pattern can be written as $\times 2+5$: $5\times 2+5=15$, $15\times 2+5=35$, $35\times 2+5=75$, $75\times 2+5=155$. Hence option (b) is correct.

Number analogy Compare the two completed statements and look for one rule that turns the pair of numbers into the result. For the pair 72 and 24: $\frac{72-24}{3}=\frac{48}{3}=16$, which is the value given. Test the same rule on the second pair: $\frac{90-30}{3}=\frac{60}{3}=20$ — it holds. Now apply it to 36 and 48: $\frac{36-48}{3}=\frac{-12}{3}=-4$. Note that the second number here is the larger one, so the result is negative. Hence option (a) is correct.

Symbol operation Find what the sign @ does by testing the first statement: $42\times 28=1176$. Since $7056÷1176=6$, the rule is (first number) $\times$ (second number) $\times 6$. Verify it on the second statement: $35\times 24\times 6=840\times 6=5040$, which matches. Apply the rule to the asked pair: $21\times 14=294$. Therefore the value is $294\times 6=1764$. Hence option (a) is correct.

Number series Write the differences between consecutive terms: 145-72=73, 252-145=107, 395-252=143. These are not constant, so take their differences in turn: 107-73=34 and 143-107=36. The second differences rise by 2, so the next one is 38 and the next gap is 143+38=181. Hence the missing term is 395+181=576. As a check, $576={24}^2$, a neat value for an SSC series. Hence option (b) is correct.

Number analogy Work on the completed pair first, since it is the only one that shows the rule fully. In 91 the two digits give 9-1=8, and $8^2=64$, which is exactly the second number. So the rule is: second number = (first digit - second digit)${}^2$. For 72: 7-2=5, and $5^2=25$. Hence option (b) is correct.

Number series Take the differences of consecutive terms: 81-48=33, 124-81=43, 177-124=53, 240-177=63. The differences themselves increase by 10 each time, so the next difference is 63+10=73. Hence the next term is 240+73=313. Hence option (c) is correct.

Odd one out Each term is a letter followed by a number, so treat the two parts separately. The letters run B, D, F, H, J — a steady jump of +2, so no letter is out of place. The numbers should be consecutive perfect squares: ${11}^2=121$, ${12}^2=144$, ${13}^2=169$, ${14}^2=196$. The fifth number must therefore be ${15}^2=225$, but the series shows 230. So J230 is the term that breaks the pattern; it should have been J225. Hence option (d) is correct.

Letter series completion The series has 15 places, of which 7 are blank — the same as the number of letters in each option. The only b that is actually printed stands at place 12, and 12 is the second place of the third block of five. Try blocks of five with b in the second place: a b a a a. Every printed letter agrees with this block. So the series is a b a a a / a b a a a / a b a a a. Reading the blanks at places 2, 4, 5, 7, 8, 11, 14 in order gives b, a, a, b, a, a, a. Hence option (c), baabaaa, is correct.

Number series Check the gaps: 15-9=6, 21-15=6, 27-21=6. The series simply adds a constant 6 each time. Hence the next term is 27+6=33. The same terms are 3 times the odd numbers — $3\times 3,3\times 5,3\times 7,3\times 9,3\times 11=33$ — which confirms the answer. Hence option (c) is correct.

Comparison ordering Turn each clue into an inequality. 'T is less than S but more than Q' gives Q<T<S. 'R is more than P' gives P<R. R is the second lowest, so exactly one person is lighter than R, and from the previous line that person can only be P. Then Q, T and S must all be heavier than R, so the complete order is P<R<Q<T<S. Counting from the heaviest end: S is first and T is second. Hence option (b) is correct.

Comparison ordering 'R is more than T but less than Q and S' gives three facts at once: T<R, R<Q and R<S. 'P is the highest' puts P at the top of all five. The extra condition says Q>S, which fixes the order of the middle three as R<S<Q. Putting everything together: T<R<S<Q<P. The mobiles priced lower than Q are T, R and S — that is 3 mobiles. Hence option (c) is correct.

Seating arrangement Draw two rows of three flats facing each other: one row of three faces north, the other three face south. R has a south-facing flat, and U's flat is next to R's, so U is in the same south row, beside R. S is diagonally opposite U, which means S is in the north row but not directly across U; so S sits at one end of the north row. Q and T face the same direction. Only one south flat is still free, so Q and T cannot both be south — they must both take north flats, and P takes the last south flat. Q's flat is not next to S, so Q takes the north flat farthest from S, and T fills the north flat in between. The north-facing flats therefore go to S, T and Q, while R, U and P face south. Among the four names offered only Q has a north-facing flat, so option (c) is correct.

Number analogy The note forbids splitting the digits, so compare the whole numbers of each completed pair. First pair: 4697-3746=951. Third pair: 7110-6159=951 — the very same jump. So the rule is: second number = first number +951. Hence the missing number is 4821+951=5772. Hence option (a) is correct.

Interchange of numbers Interchanging the numbers 4 and 5 means every 4 in an equation becomes 5 and every 5 becomes 4; the other numbers and all the signs stay as they are. Equation I becomes $5\times 7+4-6÷1$; by BODMAS, 35+4-6=33, which equals the right-hand side. Equation II becomes $4\times 6+5-7÷1$; by BODMAS, 24+5-7=22, but the right-hand side is 27. So equation I is correct after the interchange and equation II is not. Hence option (a) is correct.

Signs substitution The four $*$ signs must be replaced, in order, by +, $\times$, - and $÷$. The equation then reads $22+12\times 3-42÷7=52$. By BODMAS, carry out the multiplication and the division first: $12\times 3=36$ and $42÷7=6$. Now finish with the addition and subtraction: 22+36-6=52, which is exactly the required value. No other arrangement of the given signs balances the equation, so option (c) is correct.

Odd letter-number cluster In each cluster add the alphabetical positions of the four letters and compare the total with the number written after them. For BCDE, 2+3+4+5=14; for RSTU, 18+19+20+21=78; for MNOP, 13+14+15+16=58 — each total matches its own number. For JKLM, 10+11+12+13=46, but the cluster carries 48, so the rule breaks here. Hence option (d) is correct.

Letter-number pair analogy In every pair the two letters move 2 places back in the alphabet: $D\to B$, $F\to D$; $G\to E$, $H\to F$, and so on — this part is the same everywhere. So the difference must lie in the numbers; check whether the right number is the cube root of the left number. $\sqrt[3]{216}=6$, $\sqrt[3]{1000}=10$ and $\sqrt[3]{1728}=12$ — three pairs follow the rule exactly. But $\sqrt[3]{64}=4$, whereas the first pair shows 8, which is the square root instead. Hence option (a) is correct.

Odd number pair In each pair subtract the second number from the first and see whether the gap is the same everywhere. 9120-8540=580, 7350-6770=580 and 8460-7880=580 — three pairs share the same difference. But 9988-9398=590, which is 10 more than 580. Hence option (d) is correct.

Letter coding Compare PLAN with TPER letter by letter: $P\to T$, $L\to P$, $A\to E$, $N\to R$ — every letter moves 4 places forward. MOVE confirms the same shift: $M\to Q$, $O\to S$, $V\to Z$, $E\to I$. Apply +4 to TASK: T(20)+4=X, A(1)+4=E, S(19)+4=W, K(11)+4=O. So TASK is coded as XEWO and option (c) is correct.

Word formation EDUCATION supplies the letters E, D, U, C, A, T, I, O, N — note that N is available only once. DANCE needs D, A, N, C and E, and every one of them is present in EDUCATION. SOUND needs an S and TEACH needs an H, neither of which occurs in EDUCATION, while UNION needs two N's although only one is available. Hence option (a) is correct.

Word formation STONE needs the five letters S, T, O, N and E, each of them once. TONNES is made of T, O, N, N, E, S, so all five letters of STONE can be picked out of it. TENSES and ESTEEM contain no O at all, and NOTED contains no S, so none of them can give STONE. Hence option (d) is correct.

Letter-number cluster analogy Take the model pair DY256 : BV768. The first letter goes 2 back ($D\to B$), the second goes 3 back ($Y\to V$) and the number is multiplied by 3, since $256\times 3=768$. MQ110 : KN330 obeys exactly the same rule: $M\to K$, $Q\to N$ and $110\times 3=330$. Now apply it to TB180. First letter: T(20)-2=18, which is R. Second letter: B(2)-3 runs past A and wraps round to Y(25). Number: $180\times 3=540$, so the missing cluster is RY540. Hence option (d) is correct.

Letter-number cluster analogy In PQ52 : KL42 both letters move 5 places back ($P\to K$, $Q\to L$) and the number falls by 10, since 52-10=42. ST30 : NO20 repeats the rule: $S\to N$, $T\to O$ and 30-10=20. For UV50, first letter U(21)-5=16, which is P; second letter V(22)-5=17, which is Q. Number: 50-10=40, so the cluster is PQ40. Hence option (b) is correct.

Letter-number cluster analogy In FH72 : KM504 both letters move 5 places forward ($F\to K$, $H\to M$) and $504÷72=7$, so the number is multiplied by 7. MD48 : RI336 confirms it: $M\to R$, $D\to I$ and $48\times 7=336$. For PT60, first letter P(16)+5=21, which is U; second letter T(20)+5=25, which is Y. Number: $60\times 7=420$, so the cluster is UY420. Hence option (b) is correct.

Wrong term in series Read the three parts of every term separately — first letter, second letter and number. First letters: A(1), J(10), S(19), B(28), K(37) — a steady jump of +9, so this chain is correct. Second letters: M(13), R(18), W(23), B(28), G(33) — a steady jump of +5, so this chain is correct too. Numbers: $12\times 4=48$, $48\times 4=192$, $192\times 4=768$, so the last number must be $768\times 4=3072$. The series shows 3075 in place of 3072, so KG3075 is the wrong term. Hence option (b) is correct.